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# 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.

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