Differentiate x to a complex power?

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Discussion Overview

The discussion revolves around the differentiation of the expression \( x^{a+bi} \), where \( a \) and \( b \) are real numbers. Participants explore the implications of differentiating a variable raised to a complex power, considering both theoretical and practical aspects of the operation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the differentiation rule \( \frac{d}{dx}x^{k}=kx^{k-1} \) applies even when \( k \) is a complex number.
  • There is a suggestion to use Wolfram Alpha to verify the differentiation of \( x^{a+bi} \).
  • One participant expresses confusion about the differentiation process, questioning why only the real part of the exponent seems to be affected.
  • Another participant mentions the need to choose a branch of the logarithm when dealing with complex numbers, particularly in cases involving expressions like \( i^i \).
  • It is noted that the derivative of \( z^p \) with respect to \( z \) is \( pz^{p-1} \), although the proof is considered more complex.
  • Participants discuss the implications of assuming \( x \) is real and positive when applying the natural logarithm in the differentiation process.
  • One participant proposes representing \( a+ib \) as a constant \( z \) and rewriting the function as \( e^{z \ln(x)} \), applying the chain rule to derive the differentiation.

Areas of Agreement / Disagreement

Participants generally agree that differentiation of \( x^{a+bi} \) is possible, but there are multiple competing views regarding the implications of complex analysis and the assumptions about the variable \( x \). The discussion remains unresolved regarding the conditions under which the differentiation can be applied.

Contextual Notes

Participants highlight limitations regarding the assumptions about the variable \( x \), particularly its real-valued and positive nature when using the natural logarithm. The discussion also touches on the complexity of proofs in the context of complex differentiation.

dyn
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Is it possible to differentiate xa+bi where a and b are real ? if so what is the answer ?
 
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The rule ##\frac{d}{dx}x^{k}=kx^{k-1}## works even if ##k## is a complex number. Write "D[x^(a+b*I),x]" in Wolfram Alpha and see what comes as a result.
 
dyn said:
Is it possible to differentiate xa+bi where a and b are real ? if so what is the answer ?
Yes, it's possible.

Is this homework?
 
No , its not homework. So the answer is (a+bi)xa-1+bi ?
 
You don't say whether you intend "x" itself to be real or complex. It is more common to use "z" for a complex variable with z= x+ iy, x and y both real.
In any case, it is true that the derivative of [itex]z^p[/itex] with respect to z, z a complex variable, p a complex number, is [itex]pz^{p-1}[/itex] although the proof is harder.
 
I think you need to choose a branch of Logz in some cases: if you have ## i^i## , then you write this as ## e^ilog(i)## , and the value log(i) will depend on the branch of log that you chose.
 
Thanks for your replies. I asked the question because I saw it in a book but it seems strange that the number 1 is only subtracted from the real part of the power. Also I have never encountered such a function before.
 
dyn said:
it seems strange that the number 1 is only subtracted from the real part of the power.
Represent a+ib as the constant z and write the function as ez*ln(x). Apply the chain rule and you will get a factor of d/dx(ln(x)), which is 1/x. This is a factor of x-1. That is where subtracting 1 from the real part comes from.
 
  • #10
FactChecker said:
Represent a+ib as the constant z and write the function as ez*ln(x). Apply the chain rule and you will get a factor of d/dx(ln(x)), which is 1/x. This is a factor of x-1. That is where subtracting 1 from the real part comes from.
But how do you know x is Real valued and positive to use lnx?
 
  • #11
WWGD said:
But how do you know x is Real valued and positive to use lnx?
That's a good point. I don't know how much complex analysis is reasonable in answering the OP. But I guess if a complex exponent in the original function xa+bi is ok, then the natural log of a complex variable will be ok. In any case, I hope it makes the answer as to why only the real part of the exponent changes by -1 more intuitive.
 

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