# Differentiate x to a complex power?

## Main Question or Discussion Point

Is it possible to differentiate xa+bi where a and b are real ? if so what is the answer ?

hilbert2
Gold Member
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.

Mark44
Mentor
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 ?

hilbert2
Gold Member
^ Yes, that's correct.

HallsofIvy
Homework Helper
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 $z^p$ with respect to z, z a complex variable, p a complex number, is $pz^{p-1}$ although the proof is harder.

WWGD
Gold Member
2019 Award
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.

FactChecker
Gold Member
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.

WWGD
Gold Member
2019 Award
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?

FactChecker