How do you convert $- 9 - 9 \sqrt{2} i$ $-9 - 9sqrt2 i$ to polar form?

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How do you convert $- 9 - 9 \sqrt{2} i$ $-9 - 9sqrt2 i$ to polar form?

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Answer 1 · 2017-08-12T03:18:04

$- 9 - 9 \sqrt{2} i = 9 \sqrt{3} (cos θ + i s i n θ)$ $-9-9sqrt2i=9sqrt3(costheta+isi ntheta)$ , where $θ = {tan}^{- 1} \sqrt{2}$ $theta=tan^(-1)sqrt2$ and $θ$ $theta$ is in $Q 3$ $Q3$ .

Explanation:

If a complex number $a + b i = r cos θ + i r sin θ$ $a+bi=rcostheta+irsintheta$ , $a = r cos θ$ $a=rcostheta$ and $b = r sin θ$ $b=rsintheta$ and squaring and adding $a^{2} + b^{2} = r^{2}$ $a^2+b^2=r^2$ or $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ .

Hence, as we have $- 9 - 9 \sqrt{2} i$ $-9-9sqrt2i$ , $r = \sqrt{{(- 9)}^{2} + {(- 9 \sqrt{2})}^{2}} = \sqrt{81 + 162} = \sqrt{243} = 9 \sqrt{3}$ $r=sqrt((-9)^2+(-9sqrt2)^2)=sqrt(81+162)=sqrt243=9sqrt3$ .

Hence $cos θ = - \frac{1}{\sqrt{3}}$ $costheta=-1/sqrt3$ and $sin θ = - \sqrt{\frac{2}{3}}$ $sintheta=-sqrt(2/3)$ i.e. $θ$ $theta$ is in third quadrant and $tan θ = \sqrt{2}$ $tantheta=sqrt2$

and $- 9 - 9 \sqrt{2} i = 9 \sqrt{3} (cos θ + i s i n θ)$ $-9-9sqrt2i=9sqrt3(costheta+isi ntheta)$ , where $θ = {tan}^{- 1} \sqrt{2}$ $theta=tan^(-1)sqrt2$ and $θ$ $theta$ is in $Q 3$ $Q3$ .

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How do you convert $- 9 - 9 \sqrt{2} i$ $-9 - 9sqrt2 i$ to polar form?

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