How can complex numbers be elevated to complex powers?

Click For Summary
SUMMARY

The discussion centers on the mathematical manipulation of complex numbers raised to complex powers, specifically the expression \( z^w \) where \( z \) and \( w \) are complex numbers. The user initially attempted to simplify the expression but encountered discrepancies when comparing results from Wolfram Alpha. After identifying a mistake in the value of \( \theta \), the user successfully recalculated the expression, achieving results that closely matched the expected output. The conversation highlights the complexities of exponentiation with complex numbers and suggests using polar coordinates for clarity.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with exponential functions and logarithms in the complex plane
  • Knowledge of DeMoivre's Theorem and its limitations
  • Basic concepts of polar coordinates in complex analysis
NEXT STEPS
  • Study the properties of complex logarithms and their implications in exponentiation
  • Learn about polar representation of complex numbers and its applications
  • Explore advanced topics in complex analysis, such as analytic continuation
  • Investigate the algebraic approach to complex numbers and polynomial representations
USEFUL FOR

Mathematicians, students studying complex analysis, and anyone interested in the intricacies of complex exponentiation and its applications in various fields.

ForceBoy
Messages
47
Reaction score
6
Hello

I thought is would be fun to try a problem in which I had a complex number elevated to a complex power. To do this, I first tried to manipulate the general equation ## z^{w} ## (where ##z ## and ##w## are complex numbers) to look a bit more approachable. My work is as follows:

##z^{w}##
##z^{Re(w)+i Im(w)} ##
##z^{Re(w)}*e^{i Im(w) \ln(z)} ##

##\ln(z) = \ln|z| + \theta i ##

## z^{Re(w)}*e^{(\ln|z|+\theta i) i Im(w)} ##
## z^{Re(w)}*e^{i Im(w) \ln|z|- \theta Im(w)} ##
##z^{Re(w)}*e^{i Im(w) \ln|z|}*e^{-\theta Im(w)} ##
##z^{Re(w)} \frac{|z|^{i Im(w)}}{e^{\theta Im(w)}} ##

So
## z^{w} = z^{Re(w)} \frac{|z|^{i Im(w)}}{e^{\theta Im(w)}} ##

I now decided to test this equation out. I chose the number ## 1+ i ## and elevated it to the power of itself. Using wolframalpha I got that ## (1+i)^{(1+i)} \approx 0.273+ i0.584##
https://www.wolframalpha.com/input/?i=(1+i)^(1+i)

I then used the equation I ended up with and used the same number. In wolframalpha, ## (1+i)^{1} * \frac{\sqrt{2}^{i}}{e^{\frac{\pi}{2}}} \approx 0.125 + i0.266##.
https://www.wolframalpha.com/input/?i=(1+i)^1+((sqrt(2)^(i))/(e^(pi/2)))

I don't thing the values computed are even close enough to say that both expressions are equal so I must've messed up somewhere along the way. It would be very appreciated if someone could point out my mistake. In addition, are there any ways to elevate numbers to complex powers in a 'easier' way than just working from ##z^{w} ##? Thanks.
 
Physics news on Phys.org
Some rules for exponentials that work for real numbers don't work for complex numbers. ln(x) is not unique any more, that causes problems as soon as you raise a complex number to a complex power.
 
Thanks. I understand better why my way doesn't work. Just to be clear,

##z^{Re(w)} * e^{(\ln|z| +\theta i)i Im(w)} ##
##z^{Re(w)} * e^{iIm(w)\ln(w)-\theta Im(w)} ##

Is this where I messed up? I think that here I used the rule ## (x^{a})^{b} = x^{ab} ##, where I multiplied the constants. The article said that this wasn't allowed for complex ##a## and ##b##. Also one more question, I arrived at the same answer as I had before when I used DeMoivre's Theorem

##z^{a} = |z|^{a}(\cos(a\theta)+i\sin(a\theta))##
 
The theorem only works for real exponents, right?
 
Your mistake seems to be using the wrong value of ##\theta##. The correct value is ##\pi/4##.
 
  • Like
Likes   Reactions: ForceBoy
D'oh! I just saw that! I can't believe I didn't see this before!

Now, I plugged the correct formula into wolfram alpha and got

## (1+i)^1*\frac{\sqrt{2}^{i}}{e^{\frac{\pi}{4}}} \approx 0.274 +i0.584 ##
https://www.wolframalpha.com/input/?i=(1+i)^1+((sqrt(2)^(i))/(e^(pi/4)))

This answer is now pretty close to the value of ##(1+i)^{(1+i)} ##. This but some of the operations I did might've not worked for complex numbers. What is 'the deal' then? Is ## 1+i ## an exception to the complex exponent 'rules' found on the article linked below or something else?

fresh_42 said:
 
ForceBoy said:
What is 'the deal' then?
If you want to be sure, then work with complex numbers in polar coordinates: ##z=r\cdot e^{i\varphi}##. Otherwise, you can also work with ##z=x+ t\cdot y \in \mathbb{R}[t]/(t^2+1)##. This might look complicated for complex powers, because it is! What is a polynomial up to another polynomial? It's explained in the insight article I linked.
 
fresh_42 said:
z=x+t⋅y∈R[t]/(t^{2}+1)
I'm sorry, I don't understand what the ##R[t]/(t^{2}+1)## means.
 
  • #10
ForceBoy said:
I'm sorry, I don't understand what the ##R[t]/(t^{2}+1)## means.
It means that you can calculate with complex numbers as if they were real polynomials in one variable and always take the remainder of the division by ##t^2+1##. So e.g. ##t^3+ t^2 +2t +1 = (t+1)(t^2+1)+t \equiv t \operatorname{mod}(t^2+1)## would represent ##i##. This version of a representation of complex numbers should demonstrate where the problem is: we can easily multiply and add polynomials, but what should be ##t^t##? This requires some careful definitions and conventions how it should be understood, i.e. ##z^n## with ##z\in \mathbb{C}, n\in \mathbb{N}## is easy, ##z_1^{z_2}## is not.
 
  • #11
I see what you're saying. I just can't understand why ##t^2 +1## should be used. I know it evaluates to zero if ##t=i## but I don't understand why ##t^2 +1 ## should be used.

I would also like to thank everyone at this point for your insight and help.
 
  • #12
ForceBoy said:
I see what you're saying. I just can't understand why ##t^2 +1## should be used. I know it evaluates to zero if ##t=i## but I don't understand why ##t^2 +1 ## should be used.
This is the algebraic approach to the complex numbers. You identify all polynomials with zero, which have to be the zero as complex number. This means we start with all polynomials and identify all multiples of ##t^2+1## with ##0##. This is ##i^2+1=0## which is what defines complex numbers.
 
  • #13
I'm starting to see what you mean. I'll try the polar approach then look at this algebraic approach better. Thanks.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K