Question

One of the values of ${\left(\frac{(1+i)}{\sqrt{2}}\right)}^{\frac{2}{3}}$ is

• A. $\sqrt{3}+i$
• B. $-\mathrm{i}$
• C. $\mathrm{i}$
• D. $-\sqrt{3}+i$

Answer 1

The correct option is B

$-\mathrm{i}$

Explanation for correct option:

Simplification of complex number:

We have been given ${\left(\frac{(1+i)}{\sqrt{2}}\right)}^{\frac{2}{3}}$

On simplifying, we get:

$\begin{array}{rcl}z& =& {\left(\frac{(1+i)}{\sqrt{2}}\right)}^{\frac{2}{3}}\\ & =& {\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)}^{\frac{2}{3}}\\ & =& {\left(\cos \left(\frac{\mathrm{\pi }}{4}\right)+i\sin \left(\frac{\mathrm{\pi }}{4}\right)\right)}^{\frac{2}{3}}\left[\because sin\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}},cos\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\right]\\ & =& \left(\cos \left(8k+1\right)\frac{\mathrm{\pi }}{6}+i\sin \left(8k+1\right)\frac{\mathrm{\pi }}{6}\right)\mathbf{}{\left[\because r\left(cos\theta +isin\theta \right)\right]}^{\mathbf{n}}\mathbf{=}{\mathit{r}}^{\mathbf{n}}\left(cosn\theta +isinn\theta \right)\end{array}$

Substitute $k=1$, we get

$\begin{array}{rcl}z& =& \left(\cos \left(8+1\right)\frac{\mathrm{\pi }}{6}+i\sin \left(8+1\right)\frac{\mathrm{\pi }}{6}\right)\\ & =& \left(\cos \left(9\right)\frac{\mathrm{\pi }}{6}+i\sin \left(9\right)\frac{\mathrm{\pi }}{6}\right)\\ & =& \left(\cos \left(3\right)\frac{\mathrm{\pi }}{2}+i\sin \left(3\right)\frac{\mathrm{\pi }}{2}\right)\left[\because cos\frac{3\pi }{2}=0,sin\frac{3\pi }{2}=-1\right]\\ & =& -i\end{array}$

Hence, option(B) is correct.