## About "How to find locus of complex numbers"

How to find locus of complex numbers ?

To find the locus of given complex number, first we have to replace z by the complex number x + iy and simplify.

Let us look into some example problems to understand the concept.

Example 1 :

P represents the variable complex number z, find the locus of P if

Re (z + 1/z + i) = 1

Solution :

Let z = x + iy then

By equating the real part of the complex number to 1, we get

[x (x + 1) + y (y + 1)]/x2 + (y + 1) =  1

(x2 + x + y2 + y)/x2 + (y + 1) =  1

(x2 + y2 + x + y)/(x2 + (y2 + 2y + 1)) =  1

(x2 + y2 + x + y)/(x2 + y2 + 2y + 1) =  1

x2 + y2 + x + y  =  x2 + y2 + 2y + 1

x- x2 + y2 - y2 + x + y - 2y - 1  =  0

x - y - 1  = 0

Hence the locus of the given complex number is x - y - 1  =  0.

Example 2 :

P represents the variable complex number z, find the locus of P if

|z - 5i|  =  |z + 5i|

Solution :

Let z = x + iy then

|z - 5i|  =  |z + 5i|

|(x + iy) - 5i|  =  |(x + iy) + 5i|

|x + i(y - 5)|  =  |x + i(y + 5)|

√x2 (y - 5)2   =  √x(y + 5)2

Taking squares on both sides

x(y - 5)2   =  x(y + 5)2

x+ y2 - 10y + 25   =  x+ y2 + 10y+ 25

x+ y2 - x2 - y- 10y - 10 y + 25 - 25   =  0

-20y  =  0

y  =  0

Hence the locus of given complex number is y = 0.

Example 3 :

P represents the variable complex number z, find the locus of P if

| 2z − 3 | = 2

Solution :

Let z = x + iy then

| 2z − 3 | = 2

| 2(x + iy) − 3 | = 2

| 2x + i2y − 3 | = 2

| (2x − 3) + i 2y | = 2

(2x − 3)2 + (2y)2 = 2

Taking squares on both sides

(2x)2 - 2(2x)(3) + 32 + (2y)2 = 22

4x2 - 12x + 9 + 4y2 = 4

4x2 + 4y2- 12x + 9 - 4  =  0

4x2 + 4y2- 12x + 5  =  0

Hence the locus of given complex number is 4x2 + 4y2- 12x + 5  =  0.

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