Question

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

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

Answer 1

Answer: (B), (D)

If $p=q$, then number of solution of equation will be $p+q$
If $p=q$, then number of solution of equation will be finite

$z^p=z^{-q}$
Or
$|z{|}^p.e^{ip\theta }=|z{|}^{-q}.e^{-iq\theta }$
Or
$|z{|}^{p+q}.e^{i\theta (p+q)}=1$
Or
$|z{|}^{p+q}.e^{i\theta (p+q)}=e^{i2k\pi }$
Now if $|z|=1$
$\theta (p+q)=2k\pi$
$\theta =\frac{2k\pi }{p+q}$
Now if $p=q$ we get
$\theta =k\pi$
Total solutions will be
$p+q=2p=2q$.