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Friday, February 3, 2023

Question:

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

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

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

$r=sqrt{1^2+{frac{1}{sqrt{3}}}^2}$

$therefore r=frac{2}{sqrt{3}}$

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

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

$therefore theta=frac{pi}{6}$

$therefore(1+frac{1}{sqrt{3}})^n=[frac{2}{sqrt{3}}(cosfrac{pi}{6}+isinfrac{pi}{6})]^n$

$therefore(1+frac{1}{sqrt{3}})^n=(frac{2}{sqrt{3}})^n(cosfrac{npi}{6}+isinfrac{npi}{6})$

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, $m=0$

$therefore n=-3$

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