Thread: 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??

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]

behold:







so, in order to add (resp. divide) two complex numbers on the unit circle you simply add (resp. subtract) their arguments (angles).

i cant see the invalid equation.

can u show me again?

thanks...

Unfortunately, Latex does not seem to be working today. I have removed the Latex tags.

Originally Posted by Deveno

behold:

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

= (\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).

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??

Just what you wrote: 3ϴ= [(3π/2) + 2nπ]

Now, solve for ϴ.