Question

The region of the complex plane for which |Z-a|/|Z+a|=1 is (a is real) ? Answeris y-axis

Answer 1

$let \: \: z = x + iy \: \: \: \: given \\ \frac{ |z - a| }{ |z + a| } = 1 \: \: \: i.e. \\ |x - a + iy| = |x + a + iy| \\ \sqrt{ {(x - a)}^{2} + {y}^{2} } = \sqrt{ {(x + a)}^{2} + {y}^{2} }$
$- 2ax = 2ax \\ 4ax = 0 \\ x = 0 \: \: which \: represents \: y - axis$

Answer 2

The region of the complex plane is  y-axis.

Step-by-step explanation:

We have,

$\dfrac{[z-a]}{[z+a]}=1$

⇒ ${[z-a]}={[z+a]}$

Let z = x + iy

Where, x is called real part of complex number and

y is called imaginary part of complex number and  

⇒ ${[(x+iy)-a]}={[(x+iy)+a]}$

Using the property of complex number,

$[z_{1}+z_{2}]=[z_{1}]+[z_{2}]$

⇒${[(x+iy)]-[a]}={[(x+iy)]+[a]$

⇒ $x^2+(iy)^2-2(x)(iy)-a^{2}=x^2+(iy)^2+2(x)(iy)+a^{2}$

⇒ $x^2-y^2-(2xy)i-a^{2}=x^2-y^2+(2xy)i+a^{2}$

[∵ $i^{2}=-1$]

⇒ $-(2xy)i-(2xy)i=a^{2}+a^{2}$

⇒ $-(24xy)i=2a^{2}$, the region of the complex plane is  y-axis.

Hence, the region of the complex plane is  y-axis.