Question

The value of ${\left(1+\sqrt{3}i\right)}^4+{\left(1-\sqrt{3}i\right)}^4$ is

• A. $-16$
• B. $16$
• C. $14$
• D. $-14$

Answer 1

The correct option is A

$-16$

Explanation for the correct answer:

Let $z=r\left(\cos \theta +i\sin \theta \right)$ be the polar form of a complex number

According to De Moivre's theorem

${\mathbf{z}}^{\mathbf{n}}\mathbf{=}{\mathbf{r}}^{\mathbf{n}}\left(\cos \left(\mathrm{n\theta }\right)+\mathrm{isin}\left(\mathrm{n\theta }\right)\right)$

Let us consider,

$z=\left(1+\sqrt{3}i\right)$

$\Rightarrow$ $z=2\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)$

$\Rightarrow$ $z=2\left(\cos \frac{\mathrm{\pi }}{3}+i\sin \frac{\mathrm{\pi }}{3}\right)$ $...\left(i\right)$

Comparing $\left(i\right)$ with standard polar form of complex number we get,

$r=2$, $\theta =\frac{\mathrm{\pi }}{3}$

By applying De Moivre's theorem

${\left(1+\sqrt{3}i\right)}^4=2^4\left(\cos \frac{4\mathrm{\pi }}{3}+i\sin \frac{4\mathrm{\pi }}{3}\right)$

$=16\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)$

$=-8-8i\sqrt{3}$

Similarly we can write,

${\left(1-\sqrt{3}i\right)}^4=2^4\left(\cos \frac{4\mathrm{\pi }}{3}-i\sin \frac{4\mathrm{\pi }}{3}\right)$

$=16\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$

$=-8+8i\sqrt{3}$

$\Rightarrow$ ${\left(1+\sqrt{3}i\right)}^4+{\left(1-\sqrt{3}i\right)}^4=\left(-8-8i\sqrt{3}\right)+\left(-8+8i\sqrt{3}\right)$

$\Rightarrow$ ${\left(1+\sqrt{3}i\right)}^4+{\left(1-\sqrt{3}i\right)}^4=-16$

Hence, Option (A) is the correct answer.