Differentiate x to a complex power?

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  • #1
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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  • #2
hilbert2
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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.
 
  • #3
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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?
 
  • #4
dyn
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No , its not homework. So the answer is (a+bi)xa-1+bi ?
 
  • #5
hilbert2
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^ Yes, that's correct.
 
  • #6
HallsofIvy
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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.
 
  • #7
WWGD
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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.
 
  • #8
dyn
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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.
 
  • #9
FactChecker
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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
WWGD
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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
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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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