The derivative of a function of a complex variable

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October 3rd 2011, 09:22 AM
pomocnik89

The derivative of a function of a complex variable

Hello, I'm kinda lost with this problem: "Find a derivation of this functions:"

$a) z = \frac{exp(2-it)}{exp(2+it)}$

$b) w = (3 + i)z^3exp(3z)$

I really don't know where to start, the only thing that comes to my mind is to "replace" $z$ with $x + iy$ . Can someone help me? I would like to solve it by myself, but I need a little help.

Thank you for your time and help.
October 4th 2011, 04:35 PM
MonroeYoder

Re: The derivative of a function of a complex variable

You can use your standard derivative rules from calculus to find these derivatives. There is no need to replace $z$ , just treat $i$ as a constant and use the Quotient, Product and Chain rules
October 4th 2011, 06:32 PM
HallsofIvy

Re: The derivative of a function of a complex variable

Monroe Yoder is right. As long as you have "basic" polynomial, rational, and exponential functions, it does not matter if you have "x" taking real values, or "z" taking complex values, the "rules of differentiation" are exactly the same. For example, the derivative of " $az^3e^{3z}$ " using the product rule, is just
$a(3z^2e^{3z}+ 3z^3e^{3z})$ .
October 5th 2011, 07:53 AM
pomocnik89

Re: The derivative of a function of a complex variable

Ok, thank you for you answers. I solved it this way:

$w = (3 + i)z^3e^{3z}$
$w' = \frac{dw}{dz} = 3(3 + i)z^2e^{3z} + 3(3 + i)z^3e^{3z} = 3e^{3z}(3 + i)(z^2 + z^3)$

and

$z = \frac{e^{2-it}}{e^{2+it}} = e^{-2it}$
$z' = \frac{dz}{dt} = -2ie^{-2it} =\frac{-2i}{e^{2it}}$

Am I right? Or did I miss something?
Thanks a lot.
October 5th 2011, 03:06 PM
MonroeYoder

Re: The derivative of a function of a complex variable

Looks good to me

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