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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- Dec 21st 2012, 02:23 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?? - Dec 21st 2012, 03:10 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] - Dec 21st 2012, 03:56 AMDevenoRe: The exponential form of complex number
behold:

$\displaystyle e^{i (\theta + \phi )} = \cos(\theta + \phi ) + i \sin(\theta + \phi )$

$\displaystyle = \cos(\theta)\cos(\phi ) - \sin(\theta)\sin(\phi ) + i (\sin(\theta)\cos(\phi ) + \cos(\theta)\sin(\phi ))$

$\displaystyle = (\cos(\theta) + i \sin(\theta))(\cos(\phi ) + i \sin(\phi )) = e^{i \theta}e^{i \phi }$

so, in order to add (resp. divide) two complex numbers on the unit circle you simply add (resp. subtract) their arguments (angles). - Dec 21st 2012, 06:03 AMmous99Re: The exponential form of complex number
i cant see the invalid equation.

can u show me again?

thanks... - Dec 21st 2012, 11:41 AMHallsofIvyRe: 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.