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

Question

1467 views around the world

You can reuse this answer
Creative Commons License

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#.