How do you convert $\sqrt{3} - 3 i$ $sqrt3 - 3i$ to polar form?

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Answer 1 · 2018-03-21T15:45:12

The answer is $= \sqrt{12} (cos (- \frac{1}{3} π) + i sin (- \frac{1}{3} π))$ $=sqrt12(cos(-1/3pi)+isin(-1/3pi))$

Any complex number

$z = a + i b$ $z=a+ib$

can be converted to the polar form

$z = | z | (cos θ + i sin θ)$ $z=|z|(costheta+i sintheta)$

Where,

$cos θ = \frac{a}{| z |}$ $costheta=a/(|z|)$

and

$sin θ = \frac{b}{| z |}$ $sintheta=b/(|z|)$

Here,

$z = \sqrt{3} - 3 i$ $z=sqrt3-3i$

$| z | = \sqrt{{(\sqrt{3})}^{2} + {(- 3)}^{2}} = \sqrt{12}$ $|z|=sqrt((sqrt3)^2+(-3)^2)=sqrt12$

$cos θ = \frac{\sqrt{3}}{\sqrt{12}} = \frac{1}{2}$ $costheta=sqrt3/sqrt12=1/2$

$sin θ = - \frac{3}{\sqrt{12}} = - \frac{\sqrt{3}}{2}$ $sintheta=-3/sqrt12=-sqrt3/(2)$

Therefore,

$θ = - \frac{π}{3}$ $theta=-pi/3$ , $[mod 2pi]$

$z=sqrt12(cos(-1/3pi)+isin(-1/3pi))$

How do you convert sqrt3 - 3i√3−3isqrt3 - 3i to polar form?