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Is it possible to differentiate x

^{a+bi}where a and b are real ? if so what is the answer ?You are using an out of date browser. It may not display this or other websites correctly.

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

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Mark44

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Yes, it's possible.Is it possible to differentiate x^{a+bi}where a and b are real ? if so what is the answer ?

Is this homework?

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

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^ Yes, that's correct.

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HallsofIvy

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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.

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WWGD

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FactChecker

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Represent a+ib as the constant z and write the function as eit seems strange that the number 1 is only subtracted from the real part of the power.

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WWGD

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But how do you know x is Real valued and positive to use lnx?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.

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FactChecker

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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 xBut how do you know x is Real valued and positive to use lnx?

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