Question

If $z=x+i y=r(\cos \theta+i \sin \theta)$ then the values of $\sqrt{z}$ is equal to :

• A. $\pm \frac{1}{\sqrt{2}}(\sqrt{r+x}+i \sqrt{r-x})$ for $y \geq 0$
• B. $\pm \frac{1}{\sqrt{2}}(\sqrt{r+x}-i \sqrt{r-x})$ for $y \geq 0$
• C. $\pm \frac{1}{\sqrt{2}}(\sqrt{r+x}+i \sqrt{r-x})$ for $y \leq 0$
• D. $\pm \frac{1}{\sqrt{2}}(\sqrt{r+x}-i \sqrt{r-x})$ for $y \leq 0$

Answer 1

Answer: (A), (D)

using Demoivres $\sqrt{z}= \pm \sqrt{r}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)$; case (i) if $-\pi \leq \theta<0$
$\Rightarrow-\frac{\pi}{2} \leq \frac{\theta}{2}<0$ we have $\cos \frac{\theta}{2}>0$ and $\sin \frac{\theta}{2}<0 \Rightarrow \cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}$ and
$\sin \frac{\theta}{2}=-\sqrt{\frac{1-\cos \theta}{2}}$, if $0<\theta \leq \pi \Rightarrow 0<\frac{\theta}{2} \leq \frac{\pi}{2} \Rightarrow \cos \frac{\theta}{2}$ and $\sin \frac{\theta}{2}$ both positive.
Also use $\cos \theta=\frac{x}{r}$