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Question:

Let $\mathrm{z}=\frac{1-i \sqrt{3}}{2}, i=\sqrt{-1}$ Then the value of $21+\left(z+\frac{1}{z}\right)^{3}+\left(z^{2}+\frac{1}{z^{2}}\right)^{3}+\left(z^{3}+\frac{1}{z^{3}}\right)^{3}+\ldots+\left(z^{21}+\frac{1}{z^{21}}\right)^{3}$ is ________.

Solution:

$\mathrm{Z}=\frac{1-\sqrt{3} \mathrm{i}}{2}=\mathrm{e}^{-\mathrm{i} \frac{\pi}{3}}$

$z^{r}+\frac{1}{z^{r}}=2 \cos \left(-\frac{\pi}{3}\right) r=2 \cos \frac{r \pi}{3}$

$\Rightarrow 21+\sum_{\mathrm{r}=1}^{21}\left(\mathrm{z}^{\mathrm{r}}+\frac{1}{\mathrm{z}^{\mathrm{r}}}\right)^{3}=8\left(\cos ^{3} \frac{\mathrm{r} \pi}{3}\right)=2\left(\cos \mathrm{r} \pi+3 \cos \frac{\mathrm{r} \pi}{3}\right)$

$\Rightarrow 21+\left(\mathrm{z}+\frac{1}{2}\right)^{3}+\left(\mathrm{z}^{2}+\frac{1}{\mathrm{z}^{2}}\right)^{3}+\ldots \ldots .\left(\mathrm{z}^{21}+\frac{1}{\mathrm{z}^{21}}\right)^{3}$

$=21+\sum_{r=1}^{21}\left(z^{r}+\frac{1}{z^{r}}\right)^{3}$

$=21+\sum_{\mathrm{r}=1}^{21}\left(2 \cos \mathrm{r} \pi+6 \cos \frac{\mathrm{r} \pi}{3}\right)$

$=21-2-6$

$=13$

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