# Thread: [SOLVED] quotient rule on holomorphic functions

1. ## [SOLVED] quotient rule on holomorphic functions

Clearly, this is not fun.

If we decompose f(z) and g(z) into real valued functions u + iv and s + it, it seems like a reasonable first step to multiply the numerator and denominator by the complex conjugate of g (s - it), which yields something that looks very similar to the desired result. However, when we take the derivative of these real valued functions, the denominator is no longer $g^2$ and the numerator looks more like f' multiplied by the complex conjugate of g, minus g'f.

Any hints on better steps to take?

Thank you!

2. I suggest you first solve the chain rule problem. Then you can show that $\frac{d}{dz}\frac{1}{g(z)} = \frac{-g'(z)}{g(z)^2}$ and then you can apply the product rule to $\frac{f(z)}{g(z)} = f(z) \times \frac{1}{g(z)}$.

3. Originally Posted by Bruno J.
I suggest you first solve the chain rule problem. Then you can show that $\frac{d}{dz}\frac{1}{g(z)} = \frac{-g'(z)}{g(z)^2}$ and then you can apply the product rule to $\frac{f(z)}{g(z)} = f(z) \times \frac{1}{g(z)}$.
Thank you very much.

This question was actually ordinally before the chain rule question. Is there any other way to approach it? Thanks for your help!

4. Once again you are very welcome!

There might be a way to prove it without the chain rule, but it would just be equivalent to proving the chain rule for the specific case $F \circ g$ where $F(z)=z^{-1}$. Therefore I think it really makes more sense to do the other problem first in all its generality.