How do you write the complex number in trigonometric form #-7i#?

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Answer 1 · 2018-01-06T10:22:39

The answer is #=7(cos(-pi/2)+isin(-pi/2))=7e^(-1/2ipi)#

Any complex number #z=a+ib# can be represented as

#z=r(costheta+isintheta)#

Where,

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

#costheta=a/(||z||)#

and

#sintheta=b/||z||#

Here, we have

#z=0-7i#

#||z||=sqrt((0)^2+(-7)^2)=7#

#z=7((0/7)+(-7/7)i)#

#costheta=0# and #sin theta=-1#

Therefore,

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

So,

#z=7(cos(-pi/2)+isin(-pi/2))#