How do you convert $sqrt3 - 3i$ to polar form?

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Answer 1 · 2018-03-21T15:45:12

The answer is $=sqrt12(cos(-1/3pi)+isin(-1/3pi))$

Any complex number

$z=a+ib$

can be converted to the polar form

$z=|z|(costheta+i sintheta)$

Where,

$costheta=a/(|z|)$

and

$sintheta=b/(|z|)$

Here,

$z=sqrt3-3i$

$|z|=sqrt((sqrt3)^2+(-3)^2)=sqrt12$

$costheta=sqrt3/sqrt12=1/2$

$sintheta=-3/sqrt12=-sqrt3/(2)$

Therefore,

$theta=-pi/3$ , $[mod 2pi]$

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

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