How do you perform the operation in trigonometric form (cos((5pi)/3)+isin((5pi)/3))/(cospi+isinpi)cos(5π3)+isin(5π3)cosπ+isinπ?

1 Answer
Dec 4, 2016

(cos((5pi)/3)+isin((5pi)/3))/(cospi+isinpi)=cos((2pi)/3)+isin((2pi)/3)cos(5π3)+isin(5π3)cosπ+isinπ=cos(2π3)+isin(2π3)

= -1/2+isqrt3/212+i32

Explanation:

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

re^(itheta)reiθ

As such cos((5pi)/3)+isin((5pi)/3)=e^((5pi)/3i)cos(5π3)+isin(5π3)=e5π3i

and cospi+isinpi=e^(pii)cosπ+isinπ=eπi, and hence

(cos((5pi)/3)+isin((5pi)/3))/(cospi+isinpi)=e^((5pi)/3i)/e^(pii)cos(5π3)+isin(5π3)cosπ+isinπ=e5π3ieπi

= e^((5pi)/3i-pii)=e^((2pi)/3i)e5π3iπi=e2π3i

= cos((2pi)/3)+isin((2pi)/3)cos(2π3)+isin(2π3)

= -1/2+isqrt3/212+i32