**Question: **

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

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

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

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

\( \therefore r=\frac{2}{\sqrt{3}} \)

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

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

\( \therefore \theta=\frac{\pi}{6} \)

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

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

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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