Question

If $\omega =(\frac{z-i}{1+iz}{)}^{n}n$ integral, then $\omega$ lies on the unit circle for the unit circle for

• A. only even $n$
• B. only odd $n$
• C. only positive $n$
• D. all $n$

Answer 1

Answer: (D)

$\omega ={\left(\frac{z-1}{1+iz}\right)}^n={\left[\frac{z-i}{i\left(z-i\right)}\right]}^n$
$={\left(\frac{1}{i}\right)}^n={\left(-i\right)}^n$
$\therefore \left|\omega \right|=\left|{\left(-i\right)}^n\right|={\left|-i\right|}^n=1$ for all $n$.
$\therefore \omega$ lies on unit circle for of all $n$.