Question

If z lies on the circle $|$z$|$ = 1, then $\frac{2}{z}$ lies on

• A. + = 4
• B. - = 2
• C. - = 4
• D. - = 8

Answer 1

The correct option is A

+ = 4

Solution:
Let z = h + ik
$|$z$|$ = 1 $\Rightarrow$ $\sqrt{(}{h}^2$ + $k^2$) = 1
Or $h^2$ + $k^2$ = 1
Let x + iy = $\frac{2}{z}$
= $\frac{2}{z\overline{z}}\times \overline{z}$

$x+iy$ = $\frac{2}{h^2+k^2}\times (h-ik)$

$x$ = $\frac{2h}{h^2+k^2}$,$y$ = $\frac{-2k}{h^2+k^2}$

$x^2+y^2$ = $4\frac{(h^2+k^2)}{(h^2+k^2{)}^2}$ = 4 $\Rightarrow$ circle

Another way of solving this is

Let $\omega$ = $\frac{2}{2}$

$|\omega |$ = $|\frac{2}{z}|$

= 2(because |z| = 1)

Which is a circle of radius 2 units and centre as the origin.