Question:
Calculate the largest integer \( n < 0 \) so that \( (1+i\frac{\sqrt{3}}{3})^n \) is purely imaginary.
Answer:
\( (1+i\frac{\sqrt{3}}{3})^n \)
\( =(1+i\frac{1}{\sqrt{3}})^n \)
\( r=\sqrt{1^2+{\frac{1}{\sqrt{3}}}^2} \)
\( \therefore r=\frac{2}{\sqrt{3}} \)
\( \theta=tan^{-1}(\frac{\frac{1}{\sqrt{3}}}{1}) \)
\( \therefore \theta=tan^{-1}(\frac{1}{\sqrt{3}}) \)
\( \therefore \theta=\frac{\pi}{6} \)
\( \therefore(1+i\frac{1}{\sqrt{3}})^n=[\frac{2}{\sqrt{3}}(cos\frac{\pi}{6}+isin\frac{\pi}{6})]^n \)
\( \therefore(1+i\frac{1}{\sqrt{3}})^n=(\frac{2}{\sqrt{3}})^n(cos\frac{n\pi}{6}+isin\frac{n\pi}{6}) \)
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, \( m=0 \)
\( \therefore n=-3 \)
