Polar form of i^387?

Question

anonymous · Answer

I got -icis(3pi/2), but the book says it's just cis(3pi/2)

thesmartone · Answer

http://hotmath.com/hotmath_help/topics/polar-form-of-a-complex-number.html Hope that helps you...

confluxepic · Answer

The link should give you a big hint.

anonymous · Answer

I know how I'm supposed to do it... it's just not working. Please don't answer questions you can't answer...

mathmath333 · Answer

u can reduce the power

anonymous · Answer

I know that i^387 = -i. That's why I don't understand why the polar form is just cis(3pi/2), rather than -i cis(3pi/2)

anonymous · Answer

You will never have an i in front of the cosine term, only in front of the sine term.

anonymous · Answer

Sorry?

anonymous · Answer

When you're converting into polar form, only sinx would have an i or negative i in front of it, not the cosx term. As in you won't ever get something like -i cos(3pi/2)

anonymous · Answer

How would you solve it then? Since r = a^2 + b^2 and a = 0, then I thought that r = -i. If you draw it out then it seems like r = 0, but that's obviously wrong.

anonymous · Answer

The a^2 + b^2 doesn't include the actual i, only the coefficient of i. When we write the general form of a complex number as a + bi, the i and the b are separate. So if you do sqrt(a^2+b^2) to get r, it's only sqrt(0^2 + 1^2) = 1. So r is 1.

anonymous · Answer

Oh okay. Thanks, I missed half the class so all I got were the notes OTL

anonymous · Answer

No worries :) So yeah, if there's ever an i in your answer for a conversion, it will be in front of sinx and only sinx. Hopefully that clears things up.