# PLOT THE COMPLEX NUMBER IN THE COMPLEX PLANE

How to plot the complex number in the complex plane :

General form of a complex number is

a + ib

Here a is real part and b is imaginary part. We can convert the complex number as ordered pair. So, the point is (a, b).

Now we can plot the point in the graph.

Plot all four points in the same complex plane.

Example 1 :

1 + 2i, 3 - i, -2 + 2i, i

Solution :

Given, complex numbers are 1 + 2i, 3 - i, -2 + 2i, i

We are taking the real part of the complex number on the x-coordinate and the imaginary part on the y-coordinate.

Then, the ordered pair of the complex numbers are

 1 + 2i  =  (1, 2)3 - i  =  (3, -1) -2 + 2i  =  (-2, 2)i  =  (0, 1)

So, the four points are (1, 2), (3, -1), (-2, 2) and (0, 1).

By plotting the four points on the complex plane, we get

Example 2 :

2 - 3i, 1 + i, 3, -2 - i

Solution :

Given, complex numbers are 2 - 3i, 1 + i, 3, -2 - i

Then, the ordered pair of the complex numbers are

 2 - 3i  =  (2, -3)1 + i  =  (1, 1) 3  =  (3, 0)-2 - i  =  (-2, -1)

So, the four points are (2, -3), (1, 1), (3, 0) and (-2, -1).

By plotting the four points on the complex plane, we get

Example 3 :

Find the distance between two complex numbers z=  2 + 3i and z=  7 - 9i on the complex plane.

Solution :

Given, z1  =  2 + 3i and z2  =  7 - 9i are complex numbers.

Then, the ordered pair of the complex numbers are

z=  (2, 3) and z2  =  (7, -9)

Now, we have the two points (2, 3) and (7, -9)

By plotting the two points on the complex plane, we get

Finding the distance  :

Formula for distance,

d  =  √[(x- x1)2 + (y- y1)2]

(2, 3)----->(x1, y1)

(7, -9)----->(x2, y2)

d  =  √[(7 - 2)2 + (-9 - 3)2]

d  =  √[(5)2 + (-12)2]

d  =  √(25 + 144)

d  =  √169

d  =  13 units

So, the distance is 13 units.

Example 4 :

Find the distance and midpoint between two complex numbers z  =  3 + i and w  =  1 + 3i on the complex plane.

Solution :

Given, z  =  3 + i and w  =  1 + 3i are complex numbers.

Then, the ordered pair of the complex numbers are

z  =  (3, 1) and w  =  (1, 3)

Now, we have the two points (3, 1) and (1, 3)

By plotting the two points on the complex plane, we get

Finding the distance :

Formula for distance,

d  =  √[(x- x1)+ (y- y1)2]

(3, 1)----->(x1, y1)

(1, 3)----->(x2, y2)

d  =  √[(1 - 3)+ (3 - 1)2]

d  =  √[(-2)+ (2)2]

d  =  √(4 + 4)

d  =  √8

d  =  2√2 units

So, the distance is 2√2 units.

Finding the midpoint :

Formula for midpoint,

M  =  [(x1 + x2)/2, (y1 + y2)/2]

=  [(3 + 1)/2, (1 + 3)/2]

Midpoint  =  (2, 2)

So, the midpoint is (2, 2)

Example 5 :

Find the distance and midpoint between the complex number z  =  5 + 2i and its conjugate z-  =  5 - 2i on the complex plane.

Solution :

Given, z  =  5 + 2i and z-  =  5 - 2i are complex numbers.

Then, the ordered pair of the complex numbers are

z  =  (5, 2) and z-  =  (5, -2)

Now, we have the two points (5, 2) and (5, -2)

By plotting the two points on the complex plane, we get

Finding the distance :

Formula for distance,

d  =  √[(x- x1)+ (y- y1)2]

(5, 2)----->(x1, y1)

(5, -2)----->(x2, y2)

d  =  √[(5 - 5)+ (-2 - 2)2]

d  =  √[(-4)2]

d  =  √16

d  =  4 units

So, the distance is 4 units.

Finding the midpoint :

Formula for midpoint,

M  =  [(x+ x2)/2, (y+ y2)/2]

=  [(5 + 5)/2, (2 - 2)/2]

Midpoint  =  (5, 0)

So, the midpoint is (5, 0)

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