Question

If $z$ is non zero complex number satisfying $\overline{z}=i{z}^2$, then $z$ can be

• A. $\frac{\sqrt{3}}{2}-\frac{i}{2}$
• B. $-\frac{\sqrt{3}}{2}-\frac{i}{2}$
• C. $0$
• D. $i$

Answer 1

Answer: (A), (B), (D)

$i$

Given equation is $\overline{z}=i{z}^2$
Assuming $z=a+ib\Rightarrow \overline{z}=a-ib$
Now,
$z^2=(a+ib{)}^2=a^2-b^2+2abi$
Putting in the given equation, we get
$\Rightarrow a-ib=i(a^2-b^2+2abi)\Rightarrow a-ib=-2ab+i(a^2-b^2)\Rightarrow a(1+2b)+i(b^2-b-a^2)=0\Rightarrow a(1+2b)=0,(b^2-b-a^2)=0$
Using first equation, we get
$a(1+2b)=0\Rightarrow a=0\text{or}b=-\frac{1}{2}$

When $a=0$, then
$b^2-b-a^2=0\Rightarrow b^2-b=0\Rightarrow b=0,1$

When $b=-\frac{1}{2}$, then
$b^2-b-a^2=0\Rightarrow a^2=\frac{3}{4}\Rightarrow a=\pm \frac{\sqrt{3}}{2}$

Therefore, the possible complex numbers are
$z=0+0i,(0+i),(\frac{\sqrt{3}}{2}-\frac{i}{2}),(-\frac{\sqrt{3}}{2}-\frac{i}{2})$
As $z$ is a non zero complex number, so
$z=i,(\frac{\sqrt{3}}{2}-\frac{i}{2}),(-\frac{\sqrt{3}}{2}-\frac{i}{2})$