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.