# The derivative of a function of a complex variable

• Oct 3rd 2011, 10:22 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.
• Oct 4th 2011, 05:35 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
• Oct 4th 2011, 07:32 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
$a(3z^2e^{3z}+ 3z^3e^{3z})$.
• Oct 5th 2011, 08:53 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.
• Oct 5th 2011, 04:06 PM
MonroeYoder
Re: The derivative of a function of a complex variable
Looks good to me