How do I find the polar form of $3 \sqrt{2} - 3 \sqrt{2} i$ $3sqrt2 - 3sqrt2i$ ?

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Answer 1 · 2018-08-07T16:33:32

Polar form is $6 (cos (- \frac{π}{4}) + i sin (- \frac{π}{4}))$ $6(cos(-pi/4)+isin(-pi/4))$

First just find the absolute value of the complex number. We have it as $3 \sqrt{2} - 3 \sqrt{2} i$ $3sqrt2-3sqrt2i$ and its absolute value is

$\sqrt{{(3 \sqrt{2})}^{2} + {(- 3 \sqrt{2})}^{2}} = \sqrt{18 + 18} = \sqrt{36} = 6$ $sqrt((3sqrt2)^2+(-3sqrt2)^2)=sqrt(18+18)=sqrt36=6$

Hence number can be written as

$6 (\frac{3 \sqrt{2}}{6} - \frac{3 \sqrt{2}}{6} i)$ $6((3sqrt2)/6-(3sqrt2)/6i)$

or $6 (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i)$ $6(1/sqrt2-1/sqrt2i)$

As polar number is of the form $r (cos θ + i sin θ)$ $r(costheta+isintheta)$

we have $r = 6$ $r=6$ and $cos θ = \frac{1}{\sqrt{2}}$ $costheta=1/sqrt2$ and $sin θ = - \frac{1}{\sqrt{2}}$ $sintheta=-1/sqrt2$

As cosine is ratio is positive and sine ratio is negative,

$θ = \frac{π}{4}$ $theta=pi/4$

Hence, polar form is $6 (cos (- \frac{π}{4}) + i sin (- \frac{π}{4}))$ $6(cos(-pi/4)+isin(-pi/4))$

How do I find the polar form of 3√2−3√2i3sqrt2 - 3sqrt2i?