How do you evaluate e13π12i−e13π2i using trigonometric functions?

1 Answer
Miguel F.
Mar 22, 2016

Let us convert those complex numbers into their trigonometric forms.

Explanation:

Let us take the following equality:

eiθ=cosθ+isinθ

Thus:

e13π12i=cos(13π12)+isin(13π12)=
=−1+√32√2−i√3−12√2=−1−√3−i√3+i2√2=
=−(1+√3)+i(1−√3)2√2

In the same way:

e13π2i=cos(13π2)+isin(13π2)=
=0+i⋅1=i

We sum it:

e13π12i−e13π2i=−(1+√3)+i(1−√3)2√2−i=
=−(1+√3)+i(1−√3)−i⋅2√22√2=
=−(1+√3)+i(1−√3−2√2)2√2≈
≈−0.97−1.26i

Tip: in the same way that we can write exponentials as trigonometric functions, we can write trigonometric functions as exponentials.

cosθ=eiθ+e−iθ2

sinθ=eiθ−e−iθ2i

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