Calculating 11/3 with Complex Numbers: z1/3

[squenshl]

Homework Statement

I found when z = 1 the Taylor series expansion for z^1/3 by taking 1^1/3 = 1.
What if I was taking 1^1/3 = omega = e^{2i pi/3}

Homework Equations

The Attempt at a Solution

 

[squenshl]

Do I just take z = e^{2i pi/3} and just plug that into the formula for Taylor series.

 

[HallsofIvy]

To get what? If you put $e^{2i\pi/3}$ into the Taylor's series for $z^{1/3}$ about z= 1[/itex], you should get an infinite series that sums to 1.

If you want the Taylor's series for $z^{1/3}$ about $z= e^{2i\pi/3}$ you will need to evaluate the derivatives at $e^{2i\pi/3}$ and put them into
$$\sum_{n=0}^\infty \frac{f^{(n)}(e^{2i\pi/3})}{n!}(z- e^{2i\pi/3})^n$$.

 

[jackmell]

Homework Statement

I found when z = 1 the Taylor series expansion for z^1/3 by taking 1^1/3 = 1.
What if I was taking 1^1/3 = omega = e^{2i pi/3}

That's involves and interesting phenomenon in Complex Analysis: multi-functions. The function $f(z)=z^{1/3}$ has of course three values for each value of z. This is nicely (and analytically) represented by a Riemann Surfaces in 3D space which folds over itself three times representing the three different values of the root with a singular point at the origin. If you use $(1)^{1/3}=1$, you get a power series with radius equal to 1 that represents one of those surfaces, and if you use one of the other two, you get the power series representing one of the other surfaces. But that's a little hard to understand maybe. Try this, create the power series for each root of $(1)^{1/3}$, say 50 or so terms, and then plot each over the actual Riemann Surface of the function and see how each series "patches" over each of the three surfaces in this domain $|z-1|=1$.

Edit:

And yes, you plug in $z_0=e^{2n\pi i/3}$ for each term in the Taylor series where n=0, 1 or 2 for each surface of the function.

That quote is not correct. z_0 is still 1. See corrected post below.

 

[squenshl]

My series is (-1)^2/9 - 1/3(-1)^5/9(z-1) - 1/9(-1)^8/9(z-1)² + 5/81(-1)^2/9(z-1)³

 

[jackmell]

Squenshi, I'm sorry I made a serious mistake above. $z_0$ is still 1 but to obtain the power series for the other two roots, the value of f(z) is computed based on the definition of:

$$f(z)=z^{1/3}=r^{1/3}e^{i/3(\theta+2k\pi)}$$

So for example, suppose we wanted the power series representation for the root $f2(z)=z^{1/3}=r^{1/3}e^{i/3(\theta+2\pi)}$ centered at z=1. Then we would compute:

$$f(z)=\sum_{n=0}^{\infty}\frac{f2^{(n)}(1)}{n!}(z-1)^n$$

note how I'm now using the definition of f2(z). For example:

$$f2(1)=e^{2\pi i/3}$$.

$$f2'(1)=1/3 e^{-4\pi i/3}$$

$$f2''(1)=-2/9 e^{-10\pi i/3}$$

so that for the first three terms of this series I would get:

$$f2(z)=e^{2\pi i/3}+1/3 e^{-4\pi i/3}(z-1)-2/18 e^{-10\pi i/3}(z-1)^2$$

and so forth. I hope I did not make this a mess for you and if you like try and experiment with this Mathematica code which illustrates the series "patch" on the real component of the function's Riemann surface. Note with only 10 terms, it's still a pretty good fit.

f[z_] := z^{1/3}; 
f2[z_] := Exp[2*Pi*(I/3)]*f[z]; 
z0 = 1; 
mySeries = 
  N[Sum[((D[f2[z], {z, n}] /. z -> z0)/
      n!)*(z - z0)^n, {n, 0, 10}]]

thePartialRiemannSurface = Plot3D[Re[f2[x + I*y]], 
   {x, -2, 2}, {y, -2, 2}, 
   PlotStyle -> Opacity[0.2]]

mySeriesPlot = Plot3D[Re[mySeries] /. 
    z -> x + I*y, {x, 0.1, 1.9}, 
   {y, -0.9, 0.9}, PlotStyle -> Red]

Show[{thePartialRiemannSurface, mySeriesPlot}]

 

[squenshl]

So my series is wrong.
I put it into wolframalpha and get my series.

 

[jackmell]

So my series is wrong.
I put it into wolframalpha and get my series.

Hi. I don't understand how you got that series. What did you input to Alpha? And I've always found Mathematica's notation like (-1)^{5/9} awkward to understand. I think the way I did it above is the correct way.

 

[squenshl]

Cool.
Cheers.