Question

Let $z$ be a complex number satisfying equation $z^p={\overline{z}}^q$, where $p,qϵN$, then

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

Answer 1

Answer: (A), (C)

If $p\ne q$, then number of solutions of equation will be $p+q+1$

Given: $z^p=\overline{z^q}\Rightarrow z^p=\overline{z^q}$ $(\because qϵN)$

If $p=q$, then equation $z^p=\overline{z^q}$ has infinite number of solutions because any $zϵR$ will satisfy it.

If $p\ne q$, let $p>q$,

then $z^p=\overline{z^q}\Rightarrow |z{|}^p=|z{|}^q$

$\Rightarrow |z{|}^q(|z{|}^p-q-1)=0$

$\Rightarrow |z|=0$ or $|z|=1$

$|z|=0\Rightarrow z=0+i0$

$|z|=1\Rightarrow z=e^{i\theta }$

$\because z^p=\overline{z^q}$

$\Rightarrow e^{p\theta i}=e^{-q\theta i}$

$\Rightarrow e^{(p+q)\theta i}=1$

$\Rightarrow z=1^{\frac{1}{p+q}}$

$\therefore$ The number of solutions is $p+q+1$.

Hence, option $\left(A\right)$ and $\left(C\right)$ are correct.