Question

If z= x+ iy, |z-1|²+ | z+1|²=4 then find the location of z​

Answer 1

So we are given,

• $z=x+iy$

where $x,\ y\in\mathbb{R}$ such that,

$\longrightarrow|z-1|^2+|z+1|^2=4$

$\longrightarrow2\left(z^2+1\right)=4$

$\longrightarrow z^2+1=2$

$\longrightarrow z^2=1$

$\longrightarrow (x+iy)^2=1$

$\longrightarrow x^2-y^2+2xyi=1$

$\longrightarrow (x^2-y^2-1)+(2xy)i=0+0i$

Equating corresponding parts, we get,

• $x^2-y^2=1\quad\quad\dots(1)$

• $2xy=0\quad\quad\dots(2)$

(2) implies either x or y equals 0.

Let $x=0.$ Then (1) becomes,

$\longrightarrow -y^2=1$

No $y\in\mathbb{R}$ satisfy this equation.

Let $y=0.$ Then (1) becomes,

$\longrightarrow x^2=1$

$\longrightarrow x=\pm1$

Hence $z$ has two locations, (1, 0) and (-1, 0).