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

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

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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