HOW TO FIND LOCUS OF COMPLEX NUMBERS

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)²  =  1

(x² + x + y² + y)/x² + (y + 1)²  =  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 

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

x - y - 1  = 0

So, 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   =  √x² + (y + 5)²

Taking squares on both sides

x² + (y - 5)²   =  x² + (y + 5)²

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

x² + y^{2 -} x^{2 -} y² - 10y - 10 y + 25 - 25   =  0

-20y  =  0

y  =  0 

So, 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

So, the locus of given complex number is 

4x² + 4y²- 12x + 5  =  0

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