Question

The value of ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ is

• A. $2^{15}i$
• B. $2^{15}$
• C. $-2^{15}i$
• D. $6^5$

Answer 1

The correct option is C

$-2^{15}i$

Explanation for the correct answer:

The given expression ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ can be written as

${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}={\left(\frac{2\left({\displaystyle \frac{-1}{2}}+i{\displaystyle \frac{\sqrt{3}}{2}}\right)}{1-i}\right)}^{30}$

$\Rightarrow$ ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ $={\left(\frac{2\omega }{1-i}\right)}^{30}$ where, $\mathbf{\omega }\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}\mathbf{+}\mathbf{i}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}$ is a cube root of unity $\Rightarrow$ ${\mathit{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}$

$\Rightarrow$ ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}=\frac{2^{30}{\omega }^{30}}{{\left({\left(1-i\right)}^2\right)}^{15}}$

$\Rightarrow$ ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ $=\frac{2^{30}{\left({\omega }^3\right)}^{10}}{{\left(1-2i-1\right)}^{15}}$ $\left(\because {\left(a-b\right)}^2=a^2-2\mathrm{ab}+b^2\right)$

$\Rightarrow$${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ $=\frac{2^{30}{\left(1\right)}^{10}}{{\left(-2\right)}^{15}{i}^{15}}$ [Multiply and divide by $i$]

$\Rightarrow$ ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ $=\frac{-2^{15}\times i}{i^{16}}$ $\left(\because i^{4n}=1\right)$

$\Rightarrow$ ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}=-2^{15}i$

Hence, the value of ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ is $-2^{15}i$.

Hence, Option (C) is the correct answer.