Differentiate x to a complex power?

[dyn]

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

 

[hilbert2]

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]

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

Yes, it's possible.

Is this homework?

 

[dyn]

No , its not homework. So the answer is (a+bi)x^a-1+bi ?

 

[hilbert2]

^ Yes, that's correct.

 

[HallsofIvy]

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]

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.

 

[dyn]

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]

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 e^z*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]

Represent a+ib as the constant z and write the function as e^z*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]

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 x^a+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.