The product of the four values of $(1+i \sqrt{3})^{3 / 4}$ is
The product of the four values of $(1+i \sqrt{3})^{3 / 4}$ is
- $-8 \mathrm{i}$
- i
- $-8$
- $8$
Solution
Let $z=(1+i \sqrt{3})^{3 / 4}$
Product of four values of $z=|z|^4...(i)$
Now, $z=2^{3 / 4}\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{3 / 4}$
$\begin{aligned}
& z=2^{3 / 4}\left(e^{\left.i \frac{\pi}{3}\right)^{3 / 4}}=2^{3 / 4} e^{i \frac{\pi}{4}}\right. \\
& \Rightarrow|z|=2^{3 / 4}\left|e^{i \frac{\pi}{4}}\right| \\
& \Rightarrow|z|=2^{3 / 4} \times 1=2^{3 / 4} \\
& \Rightarrow|z|^4=2^3=8
\end{aligned}$
$\therefore \quad$ Product of four values of $z=8$.
Asked in: AP EAMCET 2023 (15 May Shift 2)
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