Question

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

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

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 $-\pi \le \theta <0$
$\Rightarrow -\frac{\pi }{2}\le \frac{\theta }{2}<0$ 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 $0<\theta \le \pi \Rightarrow 0<\frac{\theta }{2}\le \frac{\pi }{2}\Rightarrow \cos \frac{\theta }{2}$ and $\sin \frac{\theta }{2}$ both positive.
Also use $\cos \theta =\frac{x}{r}$