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

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

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

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

= $-1/2+isqrt3/2$

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