# Differentiating a complex function

How would u differentiate this function w.r.t. z - $$\frac{1}{z-2+3i}$$

U would need to split it and get partial derivatives right? Although im not sure how you'd split it into real and imaginary parts when the z and i are in the denominator?

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Hint: For any complex number $z$, $$\frac{1}{z}=\frac{\bar z}{|z|^2}$$ – Alex Becker Apr 18 '12 at 17:25
The quotient rule works for complex differentiation just as well as for real differentiation. – Florian Apr 18 '12 at 17:31
If "differentiate" means $d/dz$, you use the same formulas as you do in real calculus. Writing "u" and "U" in your message could be confusing in a math forum, where single letters are usually variables. We have enough confusion from "a" and "I" already... – GEdgar Apr 18 '12 at 17:31
Try differentiating $f(z) = \frac{1}{z} = z^{-1}$ first (same rules as $\mathbb{R}$ differentiation), then use the composition rule with $g(z) = f(z-a)$, where $a = -2+3i$. – copper.hat Apr 18 '12 at 17:36

There's only one variable with respect to which you're differentiating, so you don't need partial derivatives. The basic rules are the same as with real variables: $$\frac{d}{dx} \frac{1}{g(z)} = \frac{-g'(z)}{g(z)^2},$$ so $$\frac{d}{dx} \frac{1}{z-2+3i} = \frac{-1}{(z-2+3i)^2}.$$
The fact that every function that's differentiable in a neighborhood of a point can be expanded as a power series about that point is a novel thing differing from what happens with real variables. One of its consequences is that if $f'$ exists in a neighborhood of a point, then $f^{(n)}$ exists no matter how big the integer $n$ is, since that's how convergent power series behave. That doesn't happen with real variables.