How do you convert $sqrt3 - sqrt3i$ to polar form?

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Answer 1 · 2016-10-22T11:34:16

The polar form is $sqrt3(cos(-pi/4)+isin(-pi/4))$

Let $z=sqrt3-isqrt3$

Then the modulus of z is $∣z∣=sqrt((sqrt3)^2+ (sqrt3)^2) =sqrt6$
Rewriting z

$z=∣z∣(sqrt3-isqrt3)=sqrt6((sqrt3/sqrt6)-i(sqrt3/sqrt6))$

$=sqrt6(1/sqrt2-i/sqrt2)$

Comparing this to the standard form $z=r(costheta+isintheta)$

We see that $costheta=1/sqrt2$

and $sintheta=-1/sqrt2$
This can occur in the 4th quadrant and $theta =-pi/4$

And finally we have $z=sqrt6(cos(-pi/4)+sin(-pi/4))$

You can also write the resultas $z=sqrt6e^(-ipi/4)$

How do you convert sqrt3 - sqrt3isqrt3 - sqrt3i to polar form?