Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Hint: For any complex number $z$, $$\frac{1}{z}=\frac{\bar z}{|z|^2}$$ –  Alex Becker Apr 18 '12 at 17:25
2  
The quotient rule works for complex differentiation just as well as for real differentiation. –  Florian Apr 18 '12 at 17:31
2  
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

1 Answer 1

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.

share|improve this answer
    
Did you mean $\frac{d}{dz}$? Thanks a lot for this short course in complex differentiation. I never learned complex differentiation but now I know it follow same rules :) –  triomphe Nov 28 '13 at 19:19
    
Sorry---typo. I've fixed it. –  Michael Hardy Nov 28 '13 at 19:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.