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