How do you perform the operation in trigonometric form #(cos((5pi)/3)+isin((5pi)/3))/(cospi+isinpi)#?

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Answer 1 · 2016-12-04T09:27:53

#(cos((5pi)/3)+isin((5pi)/3))/(cospi+isinpi)=cos((2pi)/3)+isin((2pi)/3)#

= #-1/2+isqrt3/2#

A complex number in polar form such as #(rcostheta+irsintheta)# can be written in exponential form as

#re^(itheta)#

As such #cos((5pi)/3)+isin((5pi)/3)=e^((5pi)/3i)#

and #cospi+isinpi=e^(pii)#, and hence

#(cos((5pi)/3)+isin((5pi)/3))/(cospi+isinpi)=e^((5pi)/3i)/e^(pii)#

= #e^((5pi)/3i-pii)=e^((2pi)/3i)#

= #cos((2pi)/3)+isin((2pi)/3)#

= #-1/2+isqrt3/2#