How do you evaluate e13π8ie3π4i using trigonometric functions?

1 Answer
Feb 7, 2017

=12(1112+i(1+121))
= 0.3242+i0.2168, nearly. After some time, I would edit my answer myself, for bugs, if any.

Explanation:

Use

ei2π=cos2π+isin2π=1 and

eiπ=cosπ+isinπ=1

The given expression is

eiπei58πeiπei14π

=(cos(112.5o)+isin(112.5o))+(cos450isin45o))

=sin22.5o+cos45i(cos22.5osin45o)

=121cos4502+i12+1+cos4502

=12(1112+i(1+121))

=0.3244+i0.2168, nearly.