Question

Given z is a complex number with modulus
1. Then the equation $(\frac{1+ia}{1-ia}{)}^4=z$ has

• A. all roots, real and distinct
• B. two roots real and two imaginary
• C. three roots real and one imaginary
• D. one root real and three imaginary

Answer 1

Answer: (A)

Since ${\left(\frac{1+ia}{1-ia}\right)}^4=z$ and $\left|z\right|=1$
$\therefore \left(\frac{1+ia}{1-ia}\right)=cos\theta +isin\theta$
$\Rightarrow \frac{1+ia}{1-ia}={\left(cos\theta +isin\theta \right)}^{\frac{1}{4}}$
$=cos\frac{\left(2r\pi +\theta \right)}{4}+isin\frac{\left(2r\pi +\theta \right)}{4}$
where $r=0,1,2,3$
$\Rightarrow \frac{1+ia}{1-ia}=cos\alpha +isin\alpha$ where $\alpha$
$=\frac{2r\pi +\theta }{4}$
$\Rightarrow \frac{2}{2ia}=\frac{cos\alpha +isin\alpha +1}{cos\alpha +isin\alpha -1}$
$\Rightarrow \frac{1}{i.a}=icot\frac{\alpha }{2}$
$\Rightarrow a=-tan\frac{\alpha }{2}$
$\Rightarrow a=-tan\left(\frac{2r\pi +\theta }{8}\right)$ where $r=0,1,2,3$
$\Rightarrow a=-tan\frac{\pi }{8},-tan\left(\frac{\pi }{4}+\frac{\theta }{8}\right)$
$cot\frac{\pi }{8},cot\left(\frac{\pi }{4}+\frac{\theta }{8}\right)$