Question:
Calculate the largest integer $n < 0$ so that $(1+frac{sqrt{3}}{3})^n$ is purely imaginary.
Answer:
$(1+frac{sqrt{3}}{3})^n$
$=(1+frac{1}{sqrt{3}})^n$
$r=sqrt{1^2+{frac{1}{sqrt{3}}}^2}$
$therefore r=frac{2}{sqrt{3}}$
$theta=tan^{-1}(frac{frac{1}{sqrt{3}}}{1})$
$therefore theta=tan^{-1}(frac{1}{sqrt{3}})$
$therefore theta=frac{pi}{6}$
$therefore(1+frac{1}{sqrt{3}})^n=[frac{2}{sqrt{3}}(cosfrac{pi}{6}+isinfrac{pi}{6})]^n$
$therefore(1+frac{1}{sqrt{3}})^n=(frac{2}{sqrt{3}})^n(cosfrac{npi}{6}+isinfrac{npi}{6})$
Since, the number is purely imaginary,
$therefore(frac{2}{sqrt{3}})^ncosfrac{npi}{6}=0$
$therefore cosfrac{npi}{6}=0$
we know that, $costheta=0 implies theta=(2m-1)frac{pi}{2}$
$therefore frac{npi}{6}=(2m-1)frac{pi}{2}$
$therefore n=3(2m-1)$
Since, $n<0$
$therefore 3(2m-1)<0 implies 2m-1<0$
$implies m<frac{1}{2}$
Choose, $m=0$
$therefore n=-3$
