How do I find the trigonometric form of the complex number $sqrt3 -i$ ?

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Answer 1 · 2014-10-21T09:58:46

Let $z=sqrt{3}-i$ .

$|z|=sqrt{(sqrt{3})^2+(-1)^2}=sqrt{4}=2$

By factoring out $2$ ,

$z=2(sqrt{3}/2-1/2i)=r(cos theta+isin theta)$

by matching the real part and the imaginary part,

$Rightarrow {(r=2),(cos theta=sqrt{3}/2),(sin theta=-1/2):}$

$Rightarrow theta=-pi/6$

Hence,

$z=2[cos(-pi/6)+i sin(-pi/6)]$

since cosine is even and sine is odd, we can also write

$z=2[cos(pi/6)-isin(pi/6)]$

I hope that this was helpful.

How do I find the trigonometric form of the complex number sqrt3 -isqrt3 -i?