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 $n$.