Q50 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
The product of all the values of $(\sqrt{3}-i)^{2 / 5}$ is
- A$2(\sqrt{3}-i)$
- B$2(\sqrt{3}+i)$
- Correct answerC$2(1-\sqrt{3} i)$
- D$2(1+\sqrt{3} i)$
Solution
Let $z=(\sqrt{3}-i)^{2 / 5}=2^{2 / 5}\left(\frac{\sqrt{3}}{2}-\frac{1}{2} i\right)$
$\begin{aligned} & =2^{2 / 5} \cdot\left[e^{\left.i\left(2 n \pi-\frac{\pi}{6}\right)\right]^{2 / 5}}=2^{2 / 5} e^{i\left(\frac{4}{5} n \pi-\frac{\pi}{15}\right)}\right. \\ & \text { Product of all values }=\left(2^{2 / 5}\right)^{n i \Sigma\left(\frac{4}{5} n \pi-\frac{\pi}{15}\right)} e^i\end{aligned}$
$=\left(2^{2 / 5}\right)^n e^{\left(\frac{4}{5} \frac{n(n+1)}{2} \pi-\frac{n \pi}{15}\right)}$
Here power is $\frac{2}{5}$ i.e. 5 roots so total number of solutions is 5 . $\begin{aligned} & \therefore \text { Product of values }=\left(2^{2 / 5}\right)^5 \cdot e^{\left(\frac{4}{5} \cdot \frac{5(5+1)}{2} \pi-\frac{5 \pi}{15}\right)} \\ & =2^2 \cdot e^{i\left(12 \pi-\frac{\pi}{3}\right)} \\ & =4\left[\cos \left(12 \pi-\frac{\pi}{3}\right)+i \sin \left(12 \pi-\frac{\pi}{3}\right)\right] \\ & =4\left[\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right]=4\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)=2(1-\sqrt{3} i) \end{aligned}$
Here power is $\frac{2}{5}$ i.e. 5 roots so total number of solutions is 5 . $\begin{aligned} & \therefore \text { Product of values }=\left(2^{2 / 5}\right)^5 \cdot e^{\left(\frac{4}{5} \cdot \frac{5(5+1)}{2} \pi-\frac{5 \pi}{15}\right)} \\ & =2^2 \cdot e^{i\left(12 \pi-\frac{\pi}{3}\right)} \\ & =4\left[\cos \left(12 \pi-\frac{\pi}{3}\right)+i \sin \left(12 \pi-\frac{\pi}{3}\right)\right] \\ & =4\left[\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right]=4\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)=2(1-\sqrt{3} i) \end{aligned}$
Hence, the correct answer is option: 'C'.