Question

Let $z$ be a complex number satisfying equation $z^n={\left(\overline{z}\right)}^m,$ where $n,m\in N.$ Then

• A. If $n=m,$ then number of solutions of the equation will be finite
• B. If $n=m,$ then number of solutions of the equation will be infinite
• C. If $n\ne m,$ then number of solutions of the equation will be $n+m$
• D. If $n\ne m,$ then number of solutions of the equation will be $n+m+1$

Answer 1

Answer: (B), (D)

If $n\ne m,$ then number of solutions of the equation will be $n+m+1$

If $n=m$, then equation becomes $z^m={\left(\overline{z}\right)}^m$ and it has infinite solutions because any $z\in R$ will satisfy it.

If $n\ne m,$ let $n>m,$ then
$z^n={\left(\overline{z}\right)}^m$
$\Rightarrow |z{|}^n=|z{|}^m$
$\Rightarrow |z{|}^m(|z{|}^{n-m}-1)=0$
$\Rightarrow |z|=0$ or $|z|=1$
$|z|=0\Rightarrow z=0+i0$
$|z|=1\Rightarrow z=e^{i\theta }$
$\Rightarrow e^{(m+n)i\theta }=1$
$\Rightarrow z=1^{1/(m+n)}$
$\Rightarrow$ Number of solutions is $m+n+1.$