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??
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).
Unfortunately, Latex does not seem to be working today. I have removed the Latex tags.
Just what you wrote: 3ϴ= [(3π/2) + 2nπ]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.