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

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

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.

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, $m=0$ 
$therefore n=-3$ 
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