Thread: help with derivative

I need help with the following expression:

$\displaystyle d/dz (z^2+3i)$

Thank you.

Well just break it up:

$\displaystyle d/dz(z^2) + d/dz(3i)$
$\displaystyle = 2z + 0$
$\displaystyle = 2z$

In this case 3i is a constant, so the derivative is 0.

Thank you!!!
What if it were a bit more complicated?
For example,

$\displaystyle \frac{(z-3i)^2}{(z^2+4)^2}$?

Originally Posted by georgel

Thank you!!!
What if it were a bit more complicated?
For example,

$\displaystyle \frac{(z-3i)^2}{(z^2+4)^2}$?

I do not really know.
I know you can always use the definition, i.e. $\displaystyle f'(z)= \lim _{h \to 0} \frac{f(z+h)-f(z)}{h}$ where $\displaystyle f(z)=\frac{(z-3i)^2}{(z^2+4)^2}$. Of course $\displaystyle h\in \mathbb{C}$.

$\displaystyle \frac{(z-3i)^2}{(z^2+4)^2}$

If it were something like this, then you can rewrite it as so:

$\displaystyle (z-3i)^2(z^2+4)^{-2}$

From here you apply the product rule. (You'll have a chain rule to apply as well.)