How do you write the complex number in trigonometric form #-8+3i#?

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Answer 1 · 2017-08-14T09:30:16

The trigonometric form is #=2.92 (cos(159.4^@)+isin(159.4^@))=2.92e^(159.4^@i)#

Our complex number is

#z=-8+3i#

The trigonometric form is

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

If our complex number is #z=a+ib#

#r=|z|=sqrt(a^2+b^2)#

And

#costheta=a/|z|# and

#sintheta=b/|z|#

Therefore,

#|z|=sqrt((-8)^2+3^2)=sqrt(64+9)=sqrt73=2.92#

#costheta=-8/sqrt73#

#sintheta=3/sqrt73#

We are in the Quadrant #II#

#Theta=159.4^@#

The trigonometric form is

#z=2.92 (cos(159.4^@)+isin(159.4^@))=2.92e^(159.4^@i)#