How do you write the trigonometric form of #5/2(sqrt3-i)#?

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Answer 1 · 2018-01-20T09:18:22

The answer is #=5(cos(-pi/6)+isin(-pi/6))#

The trigonometric form of a complex number #z=a+ib# is

#z=|z|(costheta+isintheta)#

where

#costheta=a/|z|#

and

#sintheta=b/|z|#

Here #z=5/2(sqrt3-i)#

#|z|=5/2sqrt(3+1)=5/2*2=5#

#z=5(sqrt3/2-1/2i)#

#costheta=sqrt3/2#

#sintheta=-1/2#

#theta=-pi/6#, #[mod 2pi]#

Therefore,

The trigonometric form is

#z=5(cos(-pi/6)+isin(-pi/6))=5e^(-ipi/6)#