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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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    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 $a = -2+3i$.2012-04-18

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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}{dz} \frac{1}{g(z)} = \frac{-g'(z)}{g(z)^2}, $ so $ \frac{d}{dz} \frac{1}{z-2+3i} = \frac{-1}{(z-2+3i)^2}. $

Not only are the rules you learned in first-semester calculus the same, but their proofs are the same. This makes it highly tempting to think that when you do calculus with complex variables, it's all just the same as with real variables. But the truth of the matter is that you find yourself doing lots of things that have no counterparts at all in real variables. So how to compute derivatives is not the thing that's new.

Cauchy's integral formula relating integrals to residues at a point is a novel thing not found in real variables.

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.

Another thing that doesn't happen with real variables is that the radius of convergence of a power series is always the distance from the center to the nearest point where the function behaves badly. With real variables, the radius of convergence is somewhat mysterious; with complex variables there's that simple rule.

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    Sorry---typo. I've fixed it.2013-11-28