Question

How to convert 6e^(-ipi/3) to polar form?

Answer 1

I haven't a clue how to start, so I'd just like a push in the right direction if you could.

Answer 2

Opps, make that standard form.

Answer 3

6e^(-i*pi/3) is in the form r*e^(i*theta) where

r = 6
theta = pi/3

Answer 4

so the polar form is simply (r, theta) = (6, pi/3)

you can express this in trig form to get

r*cis(theta) = 6*cis(pi/3)

recall that cis(theta) = cos(theta) + i*sin(theta)

Answer 5

\[ae^{i \theta}=a(e^{i \theta})=a(\cos(\theta)+i \sin(\theta)) =acos(\theta)+i a \sin(\theta)\]

Answer 6

I actually thought it was already in polar form

Answer 7

doing this other way would convert to rectangular form

Answer 8

http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Forms.aspx

based on that, polar form is z = r*[cos(theta) + i*sin(theta)]

Answer 9

he refers to r*e^(i*theta) as "Exponential Form"

Answer 10

and also according to that site e^(i theta) is the exponential polar form
so it is still polar form

Answer 11

Thanks, my teacher didn't discuss this at all.

Answer 12

or maybe according to that site they are calling that form just exponential form lol