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

1 Answer
Dec 4, 2016

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

= #-1/2+isqrt3/2#

Explanation:

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#