Question:

Calculate the largest integer $n < 0$ so that $(1+\frac{\sqrt{3}}{3})^n$ is purely imaginary.

## Answer:

$(1+\frac{\sqrt{3}}{3})^n$

$=(1+\frac{1}{\sqrt{3}})^n$

$r=\sqrt{1^2+{\frac{1}{\sqrt{3}}}^2}$

$\therefore r=\frac{2}{\sqrt{3}}$

$\theta=tan^{-1}(\frac{\frac{1}{\sqrt{3}}}{1})$

$\therefore \theta=tan^{-1}(\frac{1}{\sqrt{3}})$

$\therefore \theta=\frac{\pi}{6}$

$\therefore(1+\frac{1}{\sqrt{3}})^n=[\frac{2}{\sqrt{3}}(cos\frac{\pi}{6}+isin\frac{\pi}{6})]^n$

$\therefore(1+\frac{1}{\sqrt{3}})^n=(\frac{2}{\sqrt{3}})^n(cos\frac{n\pi}{6}+isin\frac{n\pi}{6})$

Since, the number is purely imaginary,

$\therefore(\frac{2}{\sqrt{3}})^ncos\frac{n\pi}{6}=0$

$\therefore cos\frac{n\pi}{6}=0$

we know that, $cos\theta=0 \implies \theta=(2m-1)\frac{\pi}{2}$

$\therefore \frac{n\pi}{6}=(2m-1)\frac{\pi}{2}$

$\therefore n=3(2m-1)$

Since, $n<0$

$\therefore 3(2m-1)<0 \implies 2m-1<0$

$\implies m<\frac{1}{2}$

Choose, $m=0$

$\therefore n=-3$

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