How do you write #3(cos(90)+isin(90))# in rectangular form?

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Answer 1 · 2015-02-04T19:10:08

I think you are working in degrees here, so you simply multiply #3# times whatever you have inside your brackets.

In this case:

#cos(90°)=0#
#sin(90°)=1#

So you get:

#z=3(cos(90°)+isin(90°))=3*(0+i*1)=3i#

Your number is a pure imaginary.

You can also see this by plotting your complex number #z# on the complex plane:
enter image source here

hope it helps