How do you multiply $e^(( 3pi )/ 2 ) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2018-03-25T19:25:13

We have to use $color(red)("Euler's Identity")$ , that claims that

$e^(ix) = cos x + i sinx$ .

By plugging in $x= pi/2$ respectively, we get :

$e^(color(red)(pi/2)i) = cos color(red)(pi/2) + i sincolor(red)(pi/2) = i$

I am not sure whenever you meant $e^((3pi)/2i)$ or simply $e^((3pi)/2)$ . If you meant what you wrote, then you simply have :

$e^((3pi)/2) * e^(pi/2i) = ie^((3pi)/2)$

If not, then you can use some basic properties of powers in order to simply things :

$e^((3pi)/2 i) * e^(pi/2i) = e^((3pi+pi)/2i) = e^(2pii)$

Apply the formula again.

$e^(color(red)(2pi)i) = cos color(red)(2pi) + isincolor(red)(2pi) = 1$ .

How do you multiply e^(( 3pi )/ 2 ) * e^( pi/2 i ) e^(( 3pi )/ 2 ) * e^( pi/2 i ) in trigonometric form?