B Complex numbers and exponents

  • Thread starter ForceBoy
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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.
 
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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?
 

mathman

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Your mistake seems to be using the wrong value of ##\theta##. The correct value is ##\pi/4##.
 
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

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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

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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.
 
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.
 

fresh_42

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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.
 
I'm starting to see what you mean. I'll try the polar approach then look at this algebraic approach better. Thanks.
 

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