chwala said:
I can summarise as follows with your indulgence; if we are solving;
1. ##z^n=x^n## where ##x## is say an integer value i.e from ##z=x+iy## with ##y=0## and ##n## is a positive integer then our solution will be of the form,
##z=r^\frac{1}{n}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{1}{n}##
## =r^\frac{1}{n}\left[\dfrac{\cos (θ+2kπ)}{n}+\dfrac{\sin (θ+2kπ)}{n} \right]##
- First off, your formula is wrong. It's not ##\frac 1 n \cos(\theta + \dots )## etc. It's ##\cos(\frac{\theta + \dots}n)## etc. See the end of my post for De Moivre's Formula.
- If y = 0 (i.e., Im(z) = 0), then z is purely real and you don't need to go through all of this. In that case, z and ##z^n## will lie along the real axis and θ will be either 0 or ##2\pi##.
##z^n = x^n \Rightarrow z = \pm x##, with the sign depending on whether n is even or odd.
- x can be real, which includes integers and rational numbers.
chwala said:
2. ##z=(x+iy)^\frac{1}{n}## with ##n## being a positive integer then our solution will be of the form,
##z=r^\frac{1}{n}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{1}{n}##
## =r^\frac{1}{n}\left[\dfrac{\cos (θ+2kπ)}{n}+\dfrac{\sin (θ+2kπ)}{n} \right]##
As already noted for your first case, the above is incorrect. The angle is what gets divided/multiplied, not the cosine or sine of the angle.
chwala said:
@Mark44 like you said it does not really matter whether ##n## is positive or fractional index.
Probably the only variation would be on say;
3. ##z^n=x^\frac{n_1}{n_2}## where ##n_1## and ##n_2## are integers then, our solution will be of the form,
##z=r^\frac{n_1}{n_2}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{n_1}{n_2}##
## =r^\frac{n_1}{n_2}\left[\dfrac{\cos (n_1(θ+2kπ)}{n_2}+\dfrac{\sin n_1(θ+2kπ)}{n_2} \right]##
I would appreciate your input or comments of correction...cheers.
You don't need separate cases for #2 and #3.
If z = x + iy, where x and y are real numbers, then
##z^n = r^n [\cos(n(\theta + 2k\pi)) + i \sin(n(\theta + 2k\pi))] = r^n cis(n(\theta + 2k\pi)##
If n is a positive integer, you can omit the ##2k\pi## terms.
If n = 1/m, where m is a positive integer, k will be in the range 0, ..., m-1.
If n is a rational number in reduced form a/b, then ##z^n = z^{a/b} = (z^{1/b})^a##, so the above apply.
The basic idea when a complex number z is written in polar form is this:
##z = r cis(\theta) \Rightarrow z^n = r^n cis(n \theta)##
To raise a complex number to a power, raise the modulus (magnitude) to that power and multiply the argument (angle) by that power -- this is called De Moivre's Formula. See
https://en.wikipedia.org/wiki/De_Moivre's_formula.
One more thing. Your use of INDENT tags is really cluttering things up. Please limit your use of such things.