Algebraic Limits and its Properties

September 18, 2019/by admin_ct

What is limit ?

In simple words, limit is the output(y value) that a function approaches as the input(x value) approaches some value.

Limit of a function f(x) is usually written as

$\dpi{150} \mathbf{\lim_{x\rightarrow c}f(x) = L}$

And is read as limit of f(x) as x approaches to c, is L

Which can be easily understood as when x approaches to c, y approaches to L.

There are many techniques for evaluating limits. But before that we should get familiar with properties of limits.

Properties of limits

Let $\dpi{150} \mathbf{\lim_{x\rightarrow a}f(x)}$ and $\dpi{150} \mathbf{\lim_{x\rightarrow c}f(x) }$ exist and let c be a constant.

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\left [ f(x)\pm g(x) \right ]= \lim_{x\rightarrow a}f(x)\pm \lim_{x\rightarrow a}g(x))}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a} cf(x)= c\lim_{x\rightarrow a}f(x)}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\left [ f(x)g(x) \right ]=\lim_{x\rightarrow a}f(x)*\lim_{x\rightarrow a}g(x)}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\left [ \frac{f(x)}{g(x)} \right ]= \frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)} if \lim_{x\rightarrow a}g(x)\neq 0}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\left [ f(x) \right ]^{n}=\left [ \lim_{x\rightarrow a}f(x) \right ]^{n}}$ where n is a positive integer.

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\rightarrow a}f(x)}}$

$\dpi{150} \mathbf{\lim_{x\rightarrow a}c=c}$ [limit of constant is constant always]

Different methods to evaluate the algebraic limits are:

Limit by direct substitution
Limit by factorization
Limit by Rationalizing
By using some standard limits

Lets discuss them one by one.

1) Limit by direct substitution:

There are many well behaved functions such as polynomials , rational functions with non zero denominators for which we can use direct substitution to evaluate limit.

Example1: Evaluate the limit.

$\dpi{120} {\color{Red} \lim_{x\rightarrow 1}3x^{2}+4x+5}$

Solution: As this is a polynomial function, we just substitute x as 1 and solve.

$\dpi{120} \lim_{x\rightarrow 1}3x^{2}+4x+5 = 3(1)^{2}+4(1)+5 =12$

Example2 : Evaluate the limit.

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2}\frac{x^{2}-4}{x+3}}$

Solution: This is a rational function with non zero denominator.

So, $\dpi{120} \lim_{x\rightarrow 2}\frac{x^{2}-4}{x+3}=\frac{2^{2}-4}{2+3}= \frac{0}{5}=0$

2) Limit by factorization :

For this method, one should be familiar with all factoring techniques. If after substituting limit x->a into given rational function we get form 0/0 and (x-a) is a common factor of numerator and denominator both, then we cancel out the common factor and substitute limit in simplified form to get final answer.

Example3: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2}\frac{x^{2}-5x+6}{x^{2}-4}}$

Solution: When we plugin limit x=2 in this function, we get 0/0 . That means (x-2) is a common factor here so we try to factor numerator and denominator .

$\dpi{120} \lim_{x\rightarrow 2}\frac{x^{2}-5x+6}{x^{2}-4}= \lim_{x\rightarrow 2}\frac{(x-2)(x-3)}{(x-2)(x+2)} =\lim_{x\rightarrow 2}\frac{x-3}{x+2}=\frac{2-3}{2+2}=\frac{-1}{4}$

Example4: Evaluate

$\dpi{120} {\color{Red} \lim_{x\rightarrow 1}\frac{2}{1-x^{2}}+\frac{1}{x-1}}$

Solution: When we plugin limit x=1, it takes the form ∞+∞ . So we need to simplify it further to express it as 0/0.

$\dpi{120} \lim_{x\rightarrow 1}\frac{2}{1-x^{2}}+\frac{1}{x-1}=\lim_{x\rightarrow 1}\left [ \frac{2-(1+x)}{1-x^{2}} \right ]$

$\dpi{120} \Rightarrow \lim_{x\rightarrow 1}\frac{1-x}{1-x^{2}}=\lim_{x\rightarrow 1}\frac{1-x}{(1-x)(1+x)}$

$\dpi{120} \Rightarrow \lim_{x\rightarrow 1}\frac{1}{1+x}=\frac{1}{1+1}=\frac{1}{2}$

3) Limit by Rationalizing :

This method is used when either the numerator or denominator or both involve square roots and substituting the limit , expression take the form 0/0. Lets look at some examples.

Example5 : Evaluate the limit .

$\dpi{120} {\color{Red} \lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}}$

Solution: Here we rationalize the numerator.

$\dpi{120} \lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}=\lim_{x\rightarrow 0}\frac{\sqrt{2+x}-\sqrt{2}}{x}*\frac{\sqrt{2+x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{2}}$

$\dpi{120} =\lim_{x\rightarrow 0}\frac{\left (\sqrt{2+x}-\sqrt{2} \right )\left ( \sqrt{2+x}+\sqrt{2} \right )}{x\left (\sqrt{2+x}+\sqrt{2} \right )}$

$\dpi{120} =\lim_{x\rightarrow 0}\frac{\left (\sqrt{2+x} \right )^{2}-\left ( \sqrt{2} \right )^{2}}{x\left ( \sqrt{2+x}+\sqrt{2} \right )}$

$\dpi{120} =\lim_{x\rightarrow 0}\frac{2+x-2}{x\left ( \sqrt{2+x}+\sqrt{2} \right )}=\lim_{x\rightarrow 0}\frac{x}{x\left ( \sqrt{2+x}+\sqrt{2} \right )}$

$\dpi{120} =\lim_{x\rightarrow 0}\frac{1}{\sqrt{2+x}+\sqrt{2}}=\frac{1}{\sqrt{2}+\sqrt{2}}=\frac{1}{2\sqrt{2}}$

Example6: Find the limit

$\dpi{120} {\color{Red} \lim_ {x\rightarrow 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}}$

Solution: Here we rationalize both numerator and denominator.

$\dpi{120} \lim_ {x\rightarrow 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}=\lim_ {x\rightarrow 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}*\frac{1+\sqrt{8-x}}{1+\sqrt{8-x}}*\frac{4+\sqrt{9+x}}{4+\sqrt{9+x}}$

$\dpi{120} =\lim_{x\rightarrow 7}\frac{\ 4^{2}-\left ( \sqrt{9+x} \right )^{2}\left ( 1+\sqrt{8-x} \right )}{1^{2}-\left ( \sqrt{8-x} \right )^{2}\left ( 4+\sqrt{9+x} \right )}$

$\dpi{120} =\lim_{x\rightarrow 7}\frac{(16-9-x)\left ( 1+\sqrt{8-x} \right )}{(1-8+x)\left ( 4+\sqrt{9+x} \right )}$

$\dpi{120} =\lim_{x\rightarrow 7}\frac{(7-x)\left ( 1+\sqrt{8-x} \right )}{(-7+x)\left ( 4+\sqrt{9+x} \right )}=\lim_{x\rightarrow 7}\frac{-(-7+x)\left ( 1+\sqrt{8-x} \right )}{(-7+x)\left ( 4+\sqrt{9+x} \right )}$

$\dpi{120} =\lim_{x\rightarrow 7}\frac{-\left (1+\sqrt{8-x} \right )}{4+\sqrt{9+x}}= \frac{-\left (1+\sqrt{1} \right )}{4+\sqrt{16}}$

$\dpi{120} =\frac{-2}{4+4} =\frac{-1}{4}$

4) Using standard limits:

Here we use this formula:

$\dpi{150} \mathbf{\lim_{x\rightarrow a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}}$

Example7: Evaluate limit

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2}\frac{x^{10}-1024}{x-2}}$

Solution: When we plugin limit x=2, this expression takes the form 0/0 so we use standard formula.

$\dpi{120} lim_{x\rightarrow 2}\frac{x^{10}-1024}{x-2}= \lim_{x\rightarrow 2}\frac{x^{10}-2^{10}}{x-2}$

$\dpi{120} = 10*2^{10-1}=10*2^{9}= 5120$

Example8: If $\dpi{120} {\color{Red} \lim_{x\rightarrow a}\frac{x^{3}-a^{3}}{x-a}=\lim_{x\rightarrow a}\frac{x^{4}-1}{x-1}}$ then find all possible values of a.

Solution: Given that,

$\dpi{120} \lim_{x\rightarrow a}\frac{x^{3}-a^{3}}{x-a}=\lim_{x\rightarrow a}\frac{x^{4}-1}{x-1}$

Applying standard limit formula given above we get,

$\dpi{120} 3a^{3-1}=4(1)^{4-1}$

$\dpi{120} 3a^{2}=4$

$\dpi{120} a^{2}=\frac{4}{3}$

$\dpi{120} a=\pm \frac{2}{\sqrt{3}}$

Practice problems:

Evaluate the following limits.

$\dpi{120} \lim_{x\rightarrow -1}\frac{x^{3}-3x+1}{x-1}$

$\dpi{120} \lim_{x\rightarrow 5}\frac{x^{3}-125}{x^{2}-7x+10}$

$\dpi{120} \lim_{x\rightarrow 1}\frac{\sqrt{5x-4}-\sqrt{x}}{x-1}$

If $\dpi{120} \lim_{x\rightarrow a}\frac{x^{5}-a^{5}}{x-a}=405$ then find all possible values of a.

Answers:

-3/2
25
2
-3,3

Different forms of straight line equation

August 30, 2019/by admin_ct

A linear equation represents the equation of a straight line. There are many ways to represents the equation of a line. Here is a summary of all the forms and we will discuss them one by one in each section.

Different forms of straight line equation:

Name	Equation	Description
Standard form	ax+by+c=0	a,b,c are constants with a>0
Slope intercept form	Y=mx+b	m = slope b = y intercept
Point slope form	$\dpi{120} y-y_{1}=m(x-x_{1})$	$\dpi{120} (x_{1},y_{1})$ given point
Two point form	$\dpi{120} y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$	$\dpi{120} (x_{1},y_{1}) and \left ( x_{2},y_{2} \right )$ given points
Intercept form	$\dpi{120} \frac{x}{a}+\frac{y}{b}=1$	a= x intercept b= y intercept
Vertical line	x = a	Vertical line passing through(a,0)
Horizontal line	y= b	Horizontal line passing through(0,b)

Linear equation in standard form:

An equation of a line in standard form is given as,

ax+by+c=0

where a, b and c are integers with a>0.

Converting linear equations in standard form.

Example1: Write y= 5-2x in standard form.

Solution: First we move -2x to left side and it become positive.

Y+2x = 5

Then we move 5 too on the left side and rewrite it with x term in first place.

Y+2x-5 = 0

2x+y-5 = 0

Example2: Write equation y= 3x-6 in standard form.

Solution: First we move 3x to left and get,

y-3x = -6

Then we move -6 too on left side and rewrite the equation with x term in first place.

-3x +y+6 = 0

Now we need to change x term from negative to positive. For that we multiply both sides with -1.

3x-y-6 = 0

Example3. Write equation y= 4 – x in standard form.

Solution: For this equation , first we need to remove fractions. For that we multiply both sides with 2.

2y = 8-x

Then we move all terms to left.

2y-8+x = 0

Rewriting these terms to get standard form of equation we get,

X+2y-8=0

Example4. Write equation $\dpi{120} {\color{Red} y=\frac{5}{2}-\frac{3}{4}x}$ in standard form and then write the values of a,b and c.

Solution: To get rid of fractions here, we multiply both sides with 4 (lcm).

4y = 10-3x

Move all terms to left and we get,

4y+3x-10 = 0

Rewriting the terms to get standard form,

3x+4y-10 =0

Comparing it with standard form ax+by+c=0

We get a= 3, b=4, c=-10

Practice problems:

Write the following equations in standard form and then write the values of a,b and c.

Y=4x-5
Y= 6-3x
$\dpi{120} y=\frac{5}{2}x -3$
$\dpi{120} \frac{y}{3}=\frac{1}{2}x -4$

Answers:

4x-y-5 = 0 a=4,b=-1, c=-5
3x+y-6 = 0 a=3,b=1 ,c=-6
5x-2y-6 = 0 a=5 ,b=-2, c=-6
3x-2y-24=0 a=3,b=-2,c=-24

Slopes , Parallel and Perpendicular Lines

August 30, 2019/by admin_ct

Slope of Parallel and perpendicular lines.

What is Slope ?

Slope is a measure of the steepness of a line connecting two points and formula to calculate slope using two given points A(X1,Y1 ) and B(X2,Y2) is,

$\dpi{150} \mathbf{Slope(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}}$

Example1. Find slope of a line passing through two points(3,2) and (-1,5).

Solution: Using slope formula,

Slope(m) = $\dpi{120} \frac{5-2}{-1-3}= -\frac{3}{4}$

Example 2.Determine x so that slope of line passing through points(2,5) and (x,3) is 2.

Solution: Using slope formula,

slope(m) $\dpi{120} = \frac{3-5}{x-2}$

2 $\dpi{120} = \frac{3-5}{x-2}$

2(x-2) = -2

2x – 4 = -2

2x = 2 => x=1

Slope can also be given as tangent of the angle that a line makes with the positive direction of x axis in anticlockwise direction. It always lie between 0 and 180°.

Example3. Find slope of a line whose inclination to the positive direction of x axis is 150°.

Solution: slope = tan(150° )

tan(150) = tan(180-30) =-tan(30) = -1/√3

Therefore slope(m) $\dpi{120} = \frac{-1}{\sqrt{3}}$

Example4. Find the angle between x axis and line joining the points (3,-1) and (4,-2).

Solution: Let θ be the angle between the line and x axis. Then,

tanθ = slope of line

$\dpi{120} tan\theta = \frac{-2+1}{4-3}$

tanθ = -1 => θ = 135°

Example5. Find the value of x such that points (x,-1) (2,1) and (4,5) are collinear.

Solution: The points lying on same line are called collinear. We know that points lying on same line have same slope. Therefore

Slope of AB= slope of BC

Let the points are A(x,-1) , B(2,1) and C(4,5)

$\dpi{120} \frac{1+1}{2-x}=\frac{5-1}{4-2}$

$\dpi{120} \frac{2}{2-x}= \frac{4}{2}$

$\dpi{120} \frac{2}{2-x}= 2$

2 = 2(2-x)

2 = 4-2x

-2 = -2x => x=1

Angle between two lines:

Let θ be the angle between two lines having slopes m1 and m2, then

$\dpi{150} \mathbf{tan\theta =\left | \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right |}$

In case of parallel lines:

If two lines are parallel then angle between then would be 0.

$\dpi{120} tan\theta =\left | \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right |$

$\dpi{120} 0 =\left | \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right |$

$\dpi{120} 0=m_{2}-m_{1}$

$\dpi{120} m_{2}=m_{1}$

Therefore, when two lines are parallel, their slopes are same.

In case of perpendicular lines:

If two lines are perpendicular then angle between them is of 90°.

$\dpi{120} tan90 =\left | \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right |$

We know that tan90 is undefined , that means expression on right side will have denominator equal to 0.

$\dpi{120} 0= 1+m_{1}m_{2}$

$\dpi{120} -1= m_{1}m_{2}$

Therefore, when two lines are perpendicular, the product of their slopes is -1.

Example6. If A(-2,1) B(2,3) and C(-2,-4) are three points, find angle between BA and BC.

Solution: To find angle between lines BA and BC we need to find their slopes first.

Slope of line BA(m1 ) $\dpi{120} = \frac{3-1}{2+2} =\frac{1}{2}$

Slope of line BC(m2) $\dpi{120} =\frac{-4-3}{-2-2}= \frac{-7}{-4}=\frac{7}{4}$

Using formula , $\dpi{120} tan\theta =\left | \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right |$

$\dpi{120} tan\theta =\left |\frac{\frac{7}{4}-\frac{1}{2}}{1+\frac{7}{4}*\frac{1}{2}} \right |$

$\dpi{120} tan\theta = \left | \frac{\frac{10}{8}}{\frac{15}{8}} \right |= \frac{2}{3}$

$\dpi{120} \theta = tan^{-1}\left ( \frac{2}{3} \right )$

Example7. If the angle between two lines is π/4 and slope of one line is 1/2 , find slope of other line.

Solution: Plug in value of θ=π/4 and slope m1 = 1/2 into the formula,

$\dpi{120} tan\theta =\left | \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right |$

$\dpi{120} tan(\pi /4) =\left | \frac{m_{2}-\frac{1}{2}}{1+\frac{1}{2}m_{2}} \right |$

For the sake of simplicity lets assume m2 as m

$\dpi{120} 1 =\left |\frac{m-\frac{1}{2}}{1+\frac{m}{2}} \right |$

$\dpi{120} \pm 1=\frac{2m-1}{2+m}$

2+m = 2m-1 or -(2+m) = 2m-1

2+1 = 2m-m -2-m = 2m-1

3 = m -1/3 = m

Example8.Line through points A(-2,6) and B(4,8) is perpendicular to the line through the points C(8,12) and D(x,24). Find value of x.

Solution: Since the lines are given perpendicular , therefore product of their slopes would be -1.

Slope of AB * slope of CD = -1

$\dpi{120} \frac{8-6}{4+2}*\frac{24-12}{x-8}= -1$

$\dpi{120} \frac{2}{6}*\frac{12}{x-8}= -1$

$\dpi{120} \frac{4}{x-8}=-1$

4 = -1(x-8) => x=4

Example9. A ray of light passing through the point(1,2) reflects on x axis at point A and the reflected ray passes through the point (5,3). Find coordinates of point A.

Solution: lets assume PA is the incident ray and AQ is the reflected ray. AN is the normal which make angle of 90° with x axis. We know that incident ray and reflected ray make same angle with normal, lets assume both angles as θ . So

<PAN = <QAN =θ

Since A is the point on x axis so its coordinates would be (x,0) as y coordinate is always 0 on x axis. Slope of line PA is given as tan(90+θ ) and slope of line QA is given as tan(90-θ ).

Now we also find their slopes using points and slope formula.

Slope of line PA $\dpi{120} =\frac{2-0}{1-x}$

$\dpi{120} tan(90+\theta )=\frac{2}{1-x}$

$\dpi{120} -cot\theta =\frac{2}{1-x}$

$\dpi{120} cot\theta =\frac{-2}{1-x}$ ———-(i)

Slope of line QA $\dpi{120} =\frac{3-0}{5-x}$

$\dpi{120} tan(90-\theta )=\frac{3}{5-x}$

$\dpi{120} cot\theta = \frac{3}{5-x}$ ———(ii)

Comparing equations (i) and (ii) we get,

$\dpi{120} \frac{-2}{1-x}=\frac{3}{5-x}$ Cross multiplying

-2(5-x) = 3(1-x)

-10 +2x = 3-3x

5x = 13 => x = 13/5

Practice problems:

What is the value of x so that the line through (3,x) and (2,7) is parallel to line through(-1,4) and (0,6).
The slope of a line is double of the slope of other line. If tangent of the angle between them is 1/3, then find slope of other line.
Find the angle between x axis and line joining the points (3,1) and (4,2).
Using slope, show that A(4,4) B(3,5) and C(-1,-1) are the vertices of right triangle.

Answers:

9
±1, ±1/2
π/4

Slope Intercept form of a line

August 29, 2019/by admin_ct

Slope intercept form of a line Equation

The equation of a line with slope m and making y intercept as b is given as

y = mx+b

If there is no y intercept then b=0 and equation become as

y= mx

which means line is passing through origin.

Example1. Find equation of a line in slope intercept form which has slope as -1 and passing through point (0,4).

Solution: Given slope(m) = -1

Given point (0,4) lie on y axis because its x coordinate is 0. That means y intercept is b=4. So we get the equation as,

Y = mx +b

Y = (-1)x + 4 => -x+4

Example2. Find the equation of a straight line which cuts off an intercept of 5 units on negative direction of y axis and makes an angle of 120° with positive direction of x axis.

Solution: Given y intercept on negative y axis so b = -5

Given slope = tan(120)

m = tan(90+30) = -cot(30) = -√3

Substituting these values into formula y=mx+b we get,

Y = -√3x + (-5)

Y = -√3 x – 5

Slope of two parallel lines is always equal.

Example3. Find the equation of a line in slope intercept form which cuts off an intercept of length 3 on y axis and is parallel to the line joining the points (3,-2) and (1,4). Write the equation in standard form also.

Solution: Given y intercept b=3

Next we find slope of line passing through given points.

Slope(m) $\dpi{120} \frac{4-(-2)}{1-3}= \frac{6}{-2}=-3$

Slopes of two parallel lines are same. So the slope of required line would be same as -3.

Now plug in values m=-3 and b=3 into the equation y=mx+b,

Y= (-3)x+3 = -3x+3

In standard form this equation can be written as,

3x+y-3 =0.

Slope of perpendicular line is always negative reciprocal of slope of given line.

Example4. Find the equation of a line that has y intercept as 4 and is perpendicular to the line joining (2,-3) and (4,2). Write the equation in slope intercept form as well as standard form.

Solution: Given y intercept b = 4

Slope of given line(m) = $\dpi{120} \frac{2-(-3)}{4-2}=\frac{5}{2}$

Slope of required perpendicular line(m’) $\dpi{120} = \frac{-1}{m}=\frac{-1}{\frac{5}{2}}=\frac{-2}{5}$

Slope intercept form of the perpendicular line,

$\dpi{120} y=\frac{-2}{5}x +4$

To write it in standard form , multiply both sides with 5 to get rid of fractions.

5y = -2x +20

2x+5y-20 = 0

Slope of Horizontal line

If a line is parallel to x axis then its slope is 0, so the equation of a line parallel to x axis is just y=b.

Example5. Find the equation of a line parallel to x axis which intersect the y axis at -5.

Solution: Since line is parallel to x axis so its slope m=0

Given y intercept b=-5

So equation is given as, y=(0)x+(-5)

Y = -5

Slope of vertical line.

If a line is parallel to y axis (vertical line) then its slope would be undefined. The equation of a vertical line is always x = a where a can be any integer representing x intercept.

Example6. Find equation of a line parallel to y axis and having x intercept as 3.

Solution: We know that equation of a vertical line is always of the form x=a

Here, given a=3

So answer is x=3

Practice problems:

Find the equation of a straight line which cuts off an intercept of 2 units on negative direction of y axis and makes an angle of 150° with positive direction of x axis.
Find the equation of a line in slope intercept form which has y intercept as -4 and is parallel to the line joining the points (2,-5) and (1,2).
Find the equation of a line that has y intercept as 3 and is perpendicular to the line joining (3,5) and (4,2). Write the equation in slope intercept form.
Write the equation of lines with given following information:

Parallel to x axis and y intercept as -4
Parallel to y axis and x intercept as 7

Answers:

$\dpi{120} y=\frac{-1}{\sqrt{3}}x +2$
Y = -7x-4
$\dpi{120} y =\frac{1}{3}x+3$
y=-4 b. x=7

Point slope form of a line

August 29, 2019/by admin_ct

Point slope form of a line Equation

Equation of a line which passes through the point $\dpi{120} \left ( x_{1},y_{1} \right )$ and has slope ‘m’ is given as,

$\dpi{150} \mathbf{y-y_{1}=m(x-x_{1})}$

Example1. Find equation in point slope form of a line passing through (2,-3) and inclined at an angle of 135° with positive x axis.

Solution: Here slope m is given as tan(135) and X1=2, Y1=-3

m= tan135 = tan(90+45) = -cot45 = -1

Now we just plug in the values into point slope formula.

$\dpi{120} y-y_{1}=m(x-x_{1})$

y-(-3) = -1(x-2)

y+3 = -1(x+2) This is the equation in point slope form.

Example 2. Determine the equation of line through the point(-4,-3) and parallel to x axis.

Solution: As explained in previous section equation of a line parallel to x axis would be of the form y=kx. Where k is the y coordinate of given point. So answer would be y = -3.

Using other way too, we know that any line parallel to x axis is a horizontal line and slope of horizontal line is m=0. Using m=0 and given point (-4,-3) we get the equation as,

$\dpi{120} y-y_{1}=m(x-x_{1})$

y-(-3) = 0(x-(-4))

y+3 = 0(x+4)

y+3 = 0 Or y=-3

Example 3. Find the equation of perpendicular bisector of line segment joining the points A(2,3) and B(6,-5).

Solution: We know that slope of line perpendicular line is always the negative reciprocal. So first we find slope of line passing through given points.

Slope $\dpi{120} m= \frac{-5-3}{6-2}= \frac{-8}{4}=-2$

Slope of line perpendicular to AB $\dpi{120} = \frac{-1}{m}= \frac{-1}{-2}=\frac{1}{2}$

Next we need to find mid point of line AB to find equation of perpendicular bisector.

Using mid point formula, mid point p is given as,

$\dpi{120} \left ( \frac{2+6}{2},\frac{3-5}{2} \right )= (4,-1)$

Using slope m = 1/2 and point p(4,-1) we can write the equation in point slope form as,

y-(-1) =1/2 (x-4)

y+1 = (1/2) (x-4) This is point slope form.

Multiply both sides with 2 to get rid of fractions.

2(y+1) =(x-4)

After simplifying we get the equation in standard form as,

x-2y-6 = 0

Example4. The perpendicular from the origin to a line meets it at the point(-2,9). Find the equation of line.

Solution: First we find slope of perpendicular line OP which join the two points (-2,9) and origin(0,0).

Slope(m) = $\dpi{120} \frac{9-0}{-2-0}=\frac{-9}{2}$

Slope of line would be the negative reciprocal of slope of perpendicular line.

So, slope(m’) $\dpi{120} \frac{-1}{\frac{-9}{2}}= \frac{2}{9}$

Since the point (-2,9) also lie on the line AB having slope m’= 2/9 , so equation of line AB in point slope form is given as,

y-9 = (2/9)[x-(-2)]

y-9 = (2/9)(x+2) This is point slope form.

After simplifying it we get the answer in standard form as,

2x -9y +85 = 0

Example5. Find the equation of line passing through the point (0,2) making an angle 2π/3 with positive x axis. Also find equation of line parallel to it and crossing y axis at a distance 2 units below the origin.

Solution: Here slope is given by tan(2π/3 ).

Slope(m) = tan(2π/3) = -tan( π/3) = -√3

Using slope(m)= -√3 and point (0,2) we write the equation in point slope form as,

y-2 = -√3 (x-0)

y = -√3 x +2 This is slope intercept form.

A line parallel to this line cross y axis 2 units below the origin. So point is given as (0,-2). Being parallel to previous line, this line too would have slope of -√3 . Using this slope and point we get the equation in point slope form as,

y-(-2) = -√3 (x-0)

y+2 = -√3x

y = -√3x -2 This is slope intercept form.

Practice problems:

Find equation in point slope form of a line passing through (2,-3) and inclined at an angle of 45° with positive x axis.
Find the equation of perpendicular bisector of line segment joining the points A(3,4) and B(-1,2).
Find the equation of a line passing through(-3,5) and perpendicular to the line through the points (2,5) and (-3,6).
Two lines passing through the point (2,3) intersect each other at an angle of 60° . If slope of one line is 2, find the equation of other line.

Answers:

x-y+5 = 0
2x+y = 5
5x-y+20 = 0
$\dpi{120} y-3=\left ( \frac{2\pm \sqrt{3}}{1\mp 2\sqrt{3}} \right )(x-2)$

Domain and Range of different functions

August 28, 2019/by admin_ct

How to identify Domain and Range ?

Domain:

Set of x values for which a function is defined, is called domain of that function. Domain of different type of functions is found differently.

While finding domain we need to keep in mind the following important points:

1) There are some functions which are defined for all real numbers from – infinity to + infinity.

All polynomials , exponential functions, absolute value functions like f(x)=|x| and trigonometric functions ( sin(x) and cos(x)only) have domain from –∞ to +∞ .

2)Domain of logarithmic functions is (0, ∞) as log functions are never defined for 0 and negative values.

3) Denominator of rational functions can’t be zero.

4)Radical(sqrt) functions are never defined for negative values.

Range:

Range is the output of the function. In simple words range is the resulting y values which we get after substituting all possible x values. We can substitute different x values into the function to see the kind of y values we get. Are y values always positive or always negative?

Drawing a sketch always help to know about range of a function.

1.Example: Find domain and range of following functions.

f₁ = {(3,5),(-2,5),(1,3),(2,3)}

Domain: {-2,1,2,3}

Range: {3,5}

f₂= {(1,4),(2,3),(3,1),(0,2)}

Domain: {0,1,2,3}

Range:{1,2,3,4}

Example2 : find domain and range of following function.

f ={(0,5),(1,5),(0,4),(2,4),(1,4)}

Answer: Domain ={0,1,2}

Range = {4,5}

Example3:Find domain of polynomials.

$\dpi{120} {\color{Red} f(x)=x^{3}-5x^{2}+2x+1}$

Domain : (-∞,∞)

Range : (-∞,∞)

Domain and Range of polynomials is all Real Numbers.

Example4:Find domain of Rational expression.

$\dpi{120} {\color{Red} f(x)=\frac{1}{x+2}}$

Solution: Rational functions are not defined when denominator is 0.So to find domain we just set denominator =0 and solve for x. Those x values are not included in domain.

X+2=0

X = -2

Domain : Interval form = (-∞,-2)U(-2,∞)

Set form = R-{-2} where R is the set of all real numbers.

Range : This function can never achieve 0 value as its numerator is never 0. So 0 is not included in its range.

Interval form = (-∞,0)U(0,∞)

Set form = R-{0}

Example5: Find domain of ration function.

$\dpi{120} {\color{Red} f(x)=\frac{3x+5}{x^{2}-5x+4}}$

Solution: If possible write the function in fully factored form.

$\dpi{120} f(x)= \frac{3x+5}{(x-4)(x-1)}$

Set the denominator =0 and solve for x. Those x values will be excluded from its domain.

(x-4)(x-1)=0

X= 1,4

Domain: Interval form= (-∞,1)U(1,4)U(4,∞)

Set form = R- {1,4}

Example6: Find domain of given Rational function.

$\dpi{120} {\color{Red} f(x)= \sqrt{x-2}}$

Radical functions are never defined for negative values. Therefore to find domain of radical functions, we always set radicand ≥ 0 and solve for x.

x-2 ≥ 0

x ≥ 2

Domain = [2,∞)

Range: Output of a radical function is always positive so [0,∞)

Example7: Find domain and range of given function.

$\dpi{120} {\color{Red} f(x)=\frac{1}{\sqrt{1-x}}}$

In this function we have two conditions , first radical can’t be negative and second denominator can’t be 0. So we just set radicand >0

(1-x)>0

1 > x

OR x < 1

Domain = (-∞,1]

Range = (0,∞) This function can never be 0 because of constant numerator and output always remain positive.

Example8: Find domain of given log function.

$\dpi{120} {\color{Red} f(x)=\log_{5}(3x+6)}$

Solution: Log functions are defined for only non negative real numbers and range is set of all real numbers. So to find domain of log functions we set the function > 0 and solve for x.

3x+6 > 0

3x> -6

x > -2

Domain =(-2, ∞ )

Range = (-∞,∞)

Practice Worksheet:

Find domain of given functions.

1) $\dpi{120} f(x)=\frac{x-1}{x+3}$ Ans: D =(- ∞,-3)U(-3,∞)

2) $\dpi{120} f(x)=\frac{2x+9}{x^{2}-9}$ Ans:D=(-∞,-3)U(-3,3)U(3,∞)

3) $\dpi{120} f(x)=\sqrt{9-x^{2}}$ Ans :D=[-3,3]

Find domain and range of given functions.

4) $\dpi{120} f(x)=\frac{4-x}{x-4}$ Ans: D=(-∞,4)U(4,∞) R = -1

5) f(x)= 3x^2-5 Ans : D= (-∞,∞) R =[-5,∞)

6) $\dpi{120} f(x)=\log_{3}(3x+12)$ Ans : D=(-4,∞) R= (-∞,∞ )

What is a function?

August 28, 2019/by admin_ct

Function

A function relates an input to an output. Formal definition of a function is that a function relates each element of a set with exactly one element of other set.

A function has two properties:

Each element in X is related to some element in Y.
A function is single valued. It will not give back two or more results for the same input.

One to many is not allowed but many to one is allowed.

Examples:

Example: Find which of the following relations are functions?

A={(4,6),(1,4),(2,3),(1,6)} , B= {(3,5),(-2,5),(1,3),(2,3)}

Answer: Relation A is not a function because two points (1,4) and (1,6) have same x value 1. But relation B is a function as each pair has different x values.

Vertical line test :

If we draw a vertical line on the graph and it intersect the graph at more than one point then graph is not a function

Horizontal line test :

If we draw a horizontal line on the graph and it intersect the graph more than once, then function is not one-one.

If a graph satisfy both vertical and horizontal line test then this is one-one function.

What is a Relation ?

August 28, 2019/by admin_ct

What you should know already ?

Before knowing about relation we should know about ordered pair. An ordered pair consists of two objects or elements in a given fixed order like (x,y) First element is always x coordinate and second element is always y coordinate. Some other examples of ordered pair are (4,9), (-2,3) , (-6,-7) etc. All these are points. A point is always written in ordered form.

Relation:

Relation is a set of inputs and outputs ,written as ordered pair (x,y). Here x is input and y is output. Formal definition of relation is written as:

Let A and B be two sets .Then a relation R from A to B is a subset of AXB.

Thus R is a relation from A to B <=> R⊆AxB

There are many ways to represent a relation. A relation can be represented in Roster form, Set builder form, by arrow diagram or by just plotting the points.

Domain and range of a relation: Let R be a relation from set A to set B. then the set of all first components or x coordinates of ordered pairs, is called domain of R, while the set of all second components or y coordinates of all ordered pairs of R is called range of relation R.

Example:Let A={1,3,5,7} and B={2,4,6,8} and let R={(1,8),(3,6),(5,2),(1,4)} be a relation from A to B . Find its domain and range.

Answer: Domain: {1,3,5} Range: {2,4,6,8}

To check whether it is a relation or not.

Example: If A={1,2,3}, B={4,5,6} which one of them is a relation from A to B. Give reason in support of your answer. R1= {(1,4) ,(1,5),(2,5),(3,6)} R2={(2,6),(5,1),(4,2)}

Answer: R1 is a relation from A to B because R1⊆AxB But R2 is not a relation from A to B because ordered pairs (4,2) ,(5,1) ∉ AxB So R2 is not a relation from A to B.

Trigonometric (polar) form of Complex number

August 27, 2019/by admin_ct

Some Basic terms:

Before moving to trigonometric form of complex number , we should be familiar about Modulus and Argument of complex numbers.

Modulus of complex number (|z|) : It is the absolute value which represent distance of complex number z, from origin. If

z= a+ib, then

$\dpi{120} \mathbf{\left | z \right |=\sqrt{a^2+b^2}}$

Argument : For a complex number z= a+ib,

$\dpi{120} \mathbf{tan\Theta =\frac{b}{a}}$

The angle θ is called argument of complex number z and is written as arg(z). Argument of a complex number is not unique. Two arguments of a complex number differ by integer multiple of 2π .

To find argument of a complex number, we should use following algorithm.

1. Find acute angle θ such that $\dpi{120} \mathbf{\theta = tan^{-1}\left | \frac{b}{a} \right |}$

2. Determine in which quadrant the point (a,b) belongs and use the following facts.

If p(a,b) belongs to first quadrant then arg(z)=θ

If p(a,b) belongs to 2^nd quadrant then arg(z)=π-θ

If p(a,b) belongs to 3^rd quadrant then arg(z)= θ-π

If p(a,b) belongs to 4^th quadrant then arg(z)= -θ

Example1. Find the modulus and argument of following complex number.

Z= 1+i√3

Solution: Comparing the given complex number with std. form z=a+ib,

We have a=1 , b=√3

Modulus $\dpi{120} \left | z \right |=\sqrt{a^2+b^2}=\sqrt{1^{2}+\left ( \sqrt{3} \right )^{2}}=\sqrt{1+3}=\sqrt{4}=2$

Argument $\dpi{120} \theta =tan^{-1}\left | \frac{b}{a} \right |= tan^{-1}\left | \frac{\sqrt{3}}{1} \right | = \frac{\pi }{3}$

Here a= 1 > 0 and b= √3 > 0. Since point lie in first quadrant so argument would be just itself.

arg(z) =θ= π/3

Example2. Find modulus and argument of the following complex number.

$\dpi{120} {\color{Red} Z=}{\color{Red} \frac{1}{1+i}}$

Solution: First we find given complex number in standard form.

$\dpi{120} z=\frac{1}{1+i}*\frac{1-i}{1-i}= \frac{1-i}{1^{2}-i^{2}}$

$\dpi{120} z=\frac{1-i}{1+1}=\frac{1-i}{2}=\frac{1}{2}-\frac{1}{2}i$

Modulus $\dpi{120} \left | z \right |=\sqrt{\left ( \frac{1}{2} \right )^{2}+\left ( \frac{-1}{2} \right )^{2}}= \sqrt{\frac{1}{4}}=\frac{1}{\sqrt{2}}$

Argument $\dpi{120} \theta =tan^{-1}\left | \frac{b}{a} \right |= tan^{-1}\left | \frac{\frac{-1}{2}}{\frac{1}{2}} \right | = tan^{-1}|1|= \frac{\pi }{4}$

Since, a= 1/2 and b=-1/2 so point lie in 4^th quadrant. For 4^th quadrant ,

$\dpi{120} arg(z)= -\theta =-\frac{\pi }{4}$

Example 3.Find non zero integral solutions of $\dpi{120} {\color{Red} \left | 1-i \right |^{x}= 2^{x}}$

Solution: we know that $\dpi{120} \left | 1-i \right |= \sqrt{(1)^{2}+(-1)^{2}}= \sqrt{1+1}=\sqrt{2}$

So, $\dpi{120} \left ( \sqrt{2} \right )^{x}= 2^{x}$

$\dpi{120} 2^{\frac{x}{2}}=2^{x}$

x/2 = x ( For same bases, powers can be equated)

x/2 -x = 0

-x/2 = 0 => x = 0

So given equation has no solution.

Example4.Find all non zero complex numbers satisfying $\dpi{120} {\color{Red} \overline{z}=iz^{2}}$

Solution: let z= a+ib

$\dpi{120} \overline{z}=iz^{2}$

a-ib = i(a-ib )^2

$\dpi{120} a-ib=i(a^{2}+2abi+i^{2}b^{2})$

$\dpi{120} a-ib = i(a^{2}+2abi-b^{2})$

$\dpi{120} a-ib=i(a^{2}-b^{2})+2abi^{2}$

$\dpi{120} a-ib=i(a^{2}-b^{2})-2ab$

Comparing real and imaginary parts,

a= -2ab —-(i)

$\dpi{120} -b= a^{2}-b^{2}$ —(ii)

From equation (i) we get,

a+2ab =0 => a(1+2b)=0

a = 0 or 1+2b =0

a = 0 or -1/2 = b

Using a=0 into equation(ii) we get,

-b = -b^2

b^2-b =0

b(b-1) =0

b =0 , b= 1

So for a=0 we have b=0,b=1 and following complex numbers are possible.

a=0,b=0 => z= 0+0i

a=0,b=1 => z = 0+1i

Using b=-1/2 into equation(ii) we get,

$\dpi{120} a^{2}-b^{2}+b=0 \Rightarrow a^{2}-\left ( \frac{-1}{2} \right )^{2}+\left ( \frac{-1}{2} \right )=0$

$\dpi{120} a^{2}-\frac{1}{4}-\frac{1}{2}=0 \Rightarrow a^{2}-\frac{3}{4}=0$

$\dpi{120} a^{2}=\frac{3}{4} \Rightarrow a=\pm \frac{\sqrt{3}}{2}$

So for $\dpi{120} b=\frac{-1}{2} , a=\pm \frac{\sqrt{3}}{2}$ following complex numbers are possible.

$\dpi{120} z= \frac{\sqrt{3}}{2}-\frac{1}{2}i, z= \frac{-\sqrt{3}}{2}-\frac{1}{2}i$

Hence all possible complex numbers are:

$\dpi{120} z=0,i ,\frac{\sqrt{3}}{2}-\frac{1}{2}i ,\frac{-\sqrt{3}}{2}-\frac{1}{2}i$

Polar representation of complex number

Polar representation of a complex number is given as

Z= r(cosθ +isinθ ) , where r= |z| and θ= arg(z)

Euler representation of Complex number

This is also called exponential form of complex number. let r be the modulus and θ be the argument of complex number, then Euler form of complex number is given as,

$\dpi{150} \mathbf{re^{i\theta }}$

Example5. Express the following complex number in polar form.

Z = -1-i

Solution:

$\dpi{120} r=|z| = \sqrt{(-1)^{2}+(-1)^{2}}= \sqrt{2}$

$\dpi{120} \theta = tan^{-1}\left | \frac{-1}{-1} \right | = \frac{\pi }{4}$

Since point lie in 3^rd quadrant so

$\dpi{120} arg(z)= \frac{\pi }{4}-\pi =\frac{-3\pi }{4}$

Therefore polar form is given as,

Z = r(cosθ +isinθ )

$\dpi{120} z=\sqrt{2}\left [ cos\left ( \frac{-3\pi }{4} \right )+isin\left ( \frac{-3\pi }{4} \right ) \right ]$

$\dpi{120} z=\sqrt{2}\left [ cos\left ( \frac{3\pi }{4} \right )-isin\left ( \frac{3\pi }{4} \right ) \right ]$

In Euler form this complex number can be written as,

$\dpi{120} re^{i\theta }= \sqrt{2}e^{-i\frac{3\pi }{4}}$

Example6. Express the given complex number in polar form.

$\dpi{120} {\color{Red} \frac{i-1}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}}$

Solution: Let

$\dpi{120} z=\frac{i-1}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}=\frac{i-1}{\frac{1}{2}+i\frac{\sqrt{3}}{2}}$

$\dpi{120} =\frac{2(i-1)}{1+i\sqrt{3}}$

Lets convert it into standard form,

$\dpi{120} z =\frac{2(i-1)}{1+i\sqrt{3}}*\frac{1-i\sqrt{3}}{1-i\sqrt{3}}= \frac{2(i-1)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}$

$\dpi{120} =\frac{2\left [ \left ( -1+\sqrt{3}\right )+i(1+\sqrt{3}) \right ]}{1-(-3)}$

$\dpi{120} =\left ( \frac{\sqrt{3}-1}{2} \right )+i\left ( \frac{\sqrt{3}+1}{2} \right )$

Modulus $\dpi{120} |z|=\sqrt{\left ( \frac{\sqrt{3}-1}{2} \right )^{2}+\left ( \frac{\sqrt{3}+1}{2} \right )^{2}}$

$\dpi{120} |z|=\frac{1}{2}\sqrt{\left ( \sqrt{3}-1 \right )^{2}+\left ( \sqrt{3}+1 \right )^{2}}=\frac{1}{2}\sqrt{8}=\sqrt{2}$

$\dpi{120} \theta =tan^{-1}\left | \frac{b}{a} \right |= tan^{-1}\left | \frac{\frac{\sqrt{3}-1}{2}}{\frac{\sqrt{3}+1}{2}} \right |$

$\dpi{120} =tan^{-1}\left | \frac{\sqrt{3}-1}{\sqrt{3}+1} \right | =tan^{-1}\left ( tan\left ( \frac{\pi }{4}+\frac{\pi }{6} \right ) \right )=\frac{\pi }{4}+\frac{\pi }{6}= \frac{5\pi }{12}$

Since both a and b are positive so point lie in first quadrant so $\dpi{120} \theta =\frac{5\pi }{12}$

Polar form is given as…

$\dpi{120} z=\sqrt{2}\left ( cos\frac{5\pi }{12} +isin\frac{5\pi }{12} \right )$

Other way:

This problem can also be solved using Euler form. First we convert i-1 into Euler form .

We have modulus, r= √2 and argument θ=3π/4

$\dpi{120} i-1= \sqrt{2}e^{i\frac{3\pi }{4}}$

Also $\dpi{120} cos\frac{\pi }{3}+isin\frac{\pi }{3}=e^{i\frac{\pi }{3}}$

$\dpi{120} \frac{i-1}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}= \frac{\sqrt{2}e^{i\frac{3\pi }{4}}}{e^{i\frac{\pi }{3}}}=\sqrt{2}e^{i(\frac{3\pi }{4}-\frac{\pi }{3})}=\sqrt{2}e^{i\frac{5\pi }{12}}$

Which can be converted into polar form as,

$\dpi{120} \sqrt{2}e^{i\frac{5\pi }{12}}=\sqrt{2}\left ( cos\frac{5\pi }{12} +isin\frac{5\pi }{12} \right )$

Example7. Find the least positive integral value of n for which $\dpi{120} {\color{Red} \left ( \frac{1+i}{1-i} \right )^{n}}$ is real.

Solution:

$\dpi{120} \frac{1+i}{1-i}= \frac{1+i}{1-i}*\frac{1+i}{1+i}=\frac{(1+i)^{2}}{(1+i)(1-i)}$

$\dpi{120} =\frac{1+2i+i^{2}}{1^{2}-i^{2}}=\frac{1+2i-1}{1-(-1)}=\frac{2i}{2}=i$

$\dpi{120} \left ( \frac{1+i}{1-i} \right )^{n}= \left ( i \right )^{n}$

Since $\dpi{120} i^{2}=-1$ which is real, so least value of n =2 for which given expression is real

$\dpi{120} i^{n}=i^{2}\Rightarrow n=2$

Example8. If $\dpi{120} {\color{Red} x+iy=\frac{a+ib}{a-ib}}$ , then prove that $\dpi{120} {\color{Red} x^{2}+y^{2}=1}$

Solution:

$\dpi{120} x+iy=\frac{a+ib}{a-ib}*\frac{a+ib}{a+ib}=\frac{(a+ib)^{2}}{(a-ib)(a+ib)}$

$\dpi{120} =\frac{a^2+2abi+i^{2}b^{2}}{a^{2}-i^{2}b^{2}}= \frac{a^{2}+2abi-b^{2}}{a^{2}-\left (-b^{2} \right )}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}+ \frac{2ab}{a^{2}+b^{2}}i$ ——(i)

$\dpi{120} x-iy=\frac{a-ib}{a+ib}$

$\dpi{120} x-iy=\frac{a-ib}{a+ib}*\frac{a-ib}{1-ib}= \frac{a^{2}-2abi+i^{2}b^{2}}{a^{2}-(-b)^{2}}$

$\dpi{120} =\frac{a^{2}-b^{2}-2abi}{a^{2}+b^{2}}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}-\frac{2ab}{a^{2}+b^{2}}i$ —————-(ii)

Multiply equations (i) and (ii) we get,

$\dpi{120} (x+iy)(x-iy)=\left (\frac{a^{2}-b^{2}}{a^{2}+b^{2}} +i\frac{2ab}{a^{2}+b^{2}} \right )*\left ( \frac{a^{2}-b^{2}}{a^{2}+b^{2}}-\frac{2abi}{a^{2}+b^{2}} \right )$

$\dpi{120} x^{2}+y^{2}=\left ( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \right )^{2}-\left ( \frac{2ab}{a^{2}+b^{2}}i \right )^{2}$

$\dpi{120} x^{2}+y^{2}=\frac{\left ( a^{2}-b^{2} \right )^{2}}{\left ( a^{2}+b^{2} \right )^{2}}- i^{2}\frac{(2ab)^{2}}{\left ( a^{2}+b^{2} \right )^{2}}$

$\dpi{120} x^{2}+y^{2}=\frac{\left ( a^{2}-b^{2} \right )^{2}}{\left ( a^{2}+b^{2} \right )^{2}}- (-1)\frac{(2ab)^{2}}{\left ( a^{2}+b^{2} \right )^{2}}$

$\dpi{120} x^{2}+y^{2}=\frac{\left ( a^{2}-b^{2} \right )^{2}}{\left ( a^{2}+b^{2} \right )^{2}}+\frac{(2ab)^{2}}{\left ( a^{2}+b^{2} \right )^{2}}$

$\dpi{120} x^{2}+y^{2}= \frac{\left ( a^{2}-b^{2} \right )^{2}+(2ab)^{2}}{\left (a^{2}+b^{2} \right )^{2}}$

$\dpi{120} x^{2}+y^{2}=\frac{\left ( a^{2}+b^{2} \right )^{2}}{\left ( a^{2}+b^{2} \right )^{2}} =1$

$\dpi{120} x^{2}+y^{2}=1$

Practice problems:

Find modulus and argument of the following complex number.

z= -2 + i 2√3

Write the following complex numbers in polar form.

$\dpi{120} \frac{2+i6\sqrt{3}}{5+i\sqrt{3}}$

Find the smallest positive integer value of n for which $\dpi{120} \frac{\left (1+i \right )^{n}}{\left ( 1-i \right )^{n-2}}$ is a real number.
Find the real values of θ for which the complex number $\dpi{120} \frac{1+icos\theta }{1-2icos\theta }$ is purely real.

Answers:

|z|=4 , arg(z)= 2π/3
Z = 2(cos π/3 + i sinπ/3 )
n=1
$\dpi{120} \theta =2n\pi \pm \frac{\pi }{2}, n\in Z$

Operations on Complex numbers (Worksheet)

August 26, 2019/by admin_ct

A complex number is represented as a+ib where a is real part and b is imaginary part. The complex number a+ib can also be represented as ordered pair(a,b) and plotted as a point in Argand plane. In this plane horizontal axis is called real axis and vertical axis is called imaginary axis.

Addition and Subtraction of complex numbers:

To add or subtract complex numbers , we combine their real parts and imaginary parts separately. Real part is added with real part and imaginary part is combined with imaginary part.

Here are the examples:

Example1. Let Z1 = 3+4i , Z2=-5-2i then find i) Z1+Z2 ii) Z1-Z2

iii) 3Z1 -4Z2

Solution: Z1+Z2 = (3+4i)+(-5-2i)

= 3-5+4i-2i => -2+2i

ii) Z1-Z2= (3+4i)-(-5-2i)

= 3+4i+5+2i

= 8+ 6i

iii) 3Z1 -4Z2 = 3(3+4i)-4(-5-2i)

= 3(3)+3(4i)-4(-5)-4(-2i)

= 9+12i+20+8i => 29+20i

Some facts about i(iota) in complex numbers:

i = √-1
$\mathbf{i^{2}=\left ( \sqrt{-1} \right )^{2}=-1}$
$\mathbf{i^{3}= i^{2}*i=(-1)*i =-i}$
$\mathbf{i^{4}=\left ( i^{2} \right )^{2}=(-1)^{2}=1}$
$\mathbf{i^{5}= i^{4}*i= 1*i=i}$

and so on

Using these facts we can simplify even very large powers of i as shown below. It is better to break large powers as a multiple of 4 because $\mathbf{i^{4}=1}$

Example2. Simplify the following:

a) ${\color{Red} i^{243}}$ ${\color{Red} b)} {\color{Red} i^{540}}$

$i^{240+3}= i^{240}*1^{3}$ $i^{540}=\left ( i^{4} \right )^{135}$

$\left ( i^{4} \right )^{60}*i^{3}= (1)^{60}*(-i)$ $(1)^{135}=1$

1(-i) = -i

c) $\dpi{120} \mathbf{{\color{Red} i^{7}}}$

$(i)^{6}*(i)= \left ( i^{2} \right )^{3}*i$

$\left ( -1 \right )^{3}*i$

(-1) i = -i

This way any power of i can be reduced to either 1,-1 or i ,-i

Multiplication of complex numbers:

Multiplication of two complex numbers is commutative and associative.

Commutative $\mathbf{Z_{1}Z_{2}=Z_{2}Z_{1}}$
Associative $\mathbf{\left (Z_{1}Z_{2} \right )Z_{3}= Z_{1}\left ( Z_{2}Z_{3} \right )}$

Multiplication of two complex numbers is done in usual way using FOIL and then simplified to get final result.

Example3. Multiply the given complex numbers

Z1=-2+i Z2=2+3i

Solution : Z1Z2 = (-2+i)(2+3i)

= -2(2)+i(2)-2(3i)+i(3i) (using FOIL)

= -4+2i-6i+3i^2

= -4-4i+3(-1)

= -4-3-4i => -7-4i

Conjugate of a complex number:

Conjugate of a complex number is obtained by replacing i with –i. If z=a+ib is a complex number then its conjugate is represented by $\mathbf{\bar{Z}}$ and is given as $\mathbf{\overline{z}= a-ib}$

Properties of Conjugate complex number:

$\mathbf{\overline{\overline{Z}} =Z}$ (double conjugate of a complex number is number itself)
$\mathbf{\overline{Z_{1}+Z_{2}}=\overline{Z_{1}} +\overline{Z_{2}}}$
$\mathbf{\overline{Z_{1}-Z_{2}}=\overline{Z_{1}} -\overline{Z_{2}}}$
$\mathbf{\overline{Z_{1}Z_{2}}= \overline{Z}_{1}\overline{Z}_{2}}$
$\mathbf{\overline{\left ( \frac{Z_{1}}{Z_{2}} \right )}= \frac{\overline{Z_{1}}}{\overline{Z_{2}}}}$

Multiplicative inverse (reciprocal) of a complex number:

To find multiplicative inverse of a complex number we multiply the numerator and denominator by conjugate of denominator.

Example4. Find multiplicative inverse of 2+3i .

Solution: Let z= 2+3i

Multiplicative inverse would be $z^{-1}=\frac{1}{2+3i}$

To simplify it further we rationalize the denominator. That means multiply the numerator and denominator with its conjugate 2-3i.

$z^{-1}=\frac{1}{2+3i}*\frac{2-3i}{2-3i}$

$=\frac{2-3i}{(2+3i)(2-3i)}$

$=\frac{2-3i}{2^{2}-(3i)^{2}} = \frac{2-3i}{4-9i^{2}}$

$=\frac{2-3i}{4-9(-1)}= \frac{2-3i}{13}$

$=\frac{2}{13}-\frac{3i}{13}$

Division of two complex numbers:

While dividing two complex numbers we rationalize the denominator using conjugate using same process as we used for multiplicative inverse above.

Example5. Perform the operation for given complex numbers and express the result in standard form a+ib.

${\color{Red} z_{1}=(1+i)^{2}, z_{2}= 3-i, Find \frac{z_{1}}{z_{2}}}$

Solution: First we need to simplify .

$z_{1}=(1+i)^{2}$

= (1+i)(1+i) = 1+2i+i^2 = 1+2i+(-1) = 2i

$\frac{z_{1}}{z_{2}}=\frac{2i}{3-i}$

$\frac{z_{1}}{z_{2}}=\frac{2i}{3-i}*\frac{3+i}{3+i}= \frac{2i(3+i)}{(3-i)(3+i)}$

$=\frac{6i+2i^{2}}{3^{2}-i^{2}}=\frac{6i+2(-1)}{9-(-1)}$

$=\frac{6i-2}{9+1}=\frac{-2+6i}{10}$

$=\frac{-2}{10}+\frac{6i}{10}=\frac{-1}{5}+\frac{3}{5}i$

Square root of a complex number:

To find square root of a complex number we use following process.

– Set the square root of given complex number equal to x+iy

-Take square of both sides

– Simplify and set their real and imaginary parts equal to each other.

– Solve for x and y.

Example6: Find the square root of following complex number.

5+12i

Solution: Let

$\sqrt{5+12i}= x+iy$

$\left (\sqrt{5+12i} \right )^{2}= \left (x+iy \right )^{2}$

$5+12i=x^{2}+2xyi+i^{2}y^{2}$

$5+12i=x^{2}+2xyi+(-1)y^{2}$

$5+12i=x^{2}-y^{2}+2xyi$

Equating real and imaginary parts, we get

5 = x^2 -y^2 , 12= 2xy

Using identity,

$\left (x^{2}+y^{2} \right )^{2}=\left (x^{2}-y^{2} \right )^{2}+4x^{2}y^{2}$

$\left (x^{2}+y^{2} \right )^{2}=\left (5 \right )^{2}+(12)^{2}$

$\left (x^{2}+y^{2} \right )^{2}= 169$

$\left (x^{2}+y^{2} \right )^{2}=\left (x^{2}-y^{2} \right )^{2}+4x^{2}y^{2}$

$x^{2}+y^{2} = 13$ ——–(i)

We already have

$x^{2}-y^{2} = 5$ ——–(ii)

Solving(adding) these two equations, we get

$2x^{2}=18 => x^{2}=9 => x=\pm 3$

and subtracting these two equations, we get

$y^{2}=4 => y=\pm 2$

Since we have equation, 12= 2xy which is positive so both x and y should be of same sign.

Either x=3, y=2 OR x=-3 , y=-2

$\sqrt{5+12i}= \pm (3+2i)$

Example7. Find real values of x and y for which the complex numbers $\dpi{100} {\color{Red} -3+ix^{2}y}$ and ${\color{Red} x^{2}+y+4i}$ are conjugate of each other.