Question

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

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

Answer 1

Answer: (A), (C)

The correct options are
A If $p=q$, then number of solutions of equation will be infinite
C If $p\ne q$, then number of solutions of equation will be $p+q+1$

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

If $p\ne q$.

let $p>q$, then $z^p=z^{-q}$.

$\therefore |z{|}^p=|z{|}^q$

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

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

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

$|z|=1\Rightarrow z=e^{i0}$

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

$\Rightarrow z=1^{1/(p+q)}$

Therefore, the number of solutions are $p+q+1$.