z = (-i) ^1/3

z^3 = -i

let z = re ^i ϴ

z^3 = (r^3)(e^i3ϴ) - i = e ^i[(3π/2) + 2nπ]

pls help me to understand how to find "3ϴ"?

what is the formula to find 3ϴ??

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- December 21st 2012, 03:23 AMmous99The exponential form of complex number
z = (-i) ^1/3

z^3 = -i

let z = re ^i ϴ

z^3 = (r^3)(e^i3ϴ) - i = e ^i[(3π/2) + 2nπ]

pls help me to understand how to find "3ϴ"?

what is the formula to find 3ϴ?? - December 21st 2012, 04:10 AMchiroRe: The exponential form of complex number
Hey mous99.

For this problem do you know how to use De-Moivre's formula for the angle? (Hint: its in the form of (ϴ + 2kπ)/n where n = 3 k = 0,1,2) [In this case theta = [3/2]*π] - December 21st 2012, 04:56 AMDevenoRe: The exponential form of complex number
behold:

so, in order to add (resp. divide) two complex numbers on the unit circle you simply add (resp. subtract) their arguments (angles). - December 21st 2012, 07:03 AMmous99Re: The exponential form of complex number
i cant see the invalid equation.

can u show me again?

thanks... - December 21st 2012, 12:41 PMHallsofIvyRe: The exponential form of complex number
Unfortunately, Latex does not seem to be working today. I have removed the Latex tags.

Quote:

The exponential form of complex number

z = (-i) ^1/3

z^3 = -i

let z = re ^i ϴ

z^3 = (r^3)(e^i3ϴ) - i = e ^i[(3π/2) + 2nπ]

pls help me to understand how to find "3ϴ"?

what is the formula to find 3ϴ??

Now, solve for ϴ.