When derivates w/ resp to complex variables differ from real derivatives?

[pellman]

If z is complex, the following rules are true, right?

\frac{d}{dz}z^n = nz^{n-1}

\frac{d}{dz}\ln{g(z)} = \frac{1}{g(z)} \frac{d}{dz}g(z)

\frac{d}{dz}e^{g(z)}=e^{g(z)}\frac{d}{dz}g(z)

These are of course the same rules as for real variables.

When do I need to be careful about taking derivatives with respect to complex variables?

Do all of the same rules apply as with real variables? Or do some rules fail?

 

[losiu99]

For analytic functions (defined as power series, like sine or exponential) taking derivative requires no more caution than in real case, and the "real" rules apply, thanks to the first formula you wrote and uniform convergence of power series. In general, however, complex derivative has more subtle existence issues. For example, |z|^2 is not differentiable anywhere except for zero.

 

[Landau]

Do all of the same rules apply as with real variables? Or do some rules fail?

The sum, product, quotient, and chain rule are still valid.

 

[pellman]

Ok. I rarely work with a non-analytic function so as long as its analytic I can just treat them like real numbers with regard to differentiation. Thanks! It's been a long time since I thought about this stuff.

 

[snipez90]

There are 3 cases to consider besides real functions of a real variable. The easier one involves complex functions of a real variable, which can be reduced to real functions of a real variable. For real functions of a complex variable, the derivative at a point is either 0 or it does not exist. Finally, the existence of the derivative of a complex function of a complex variable implies some rather strong conditions, namely http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations" . The derivations required in the last two cases both involve the same method, which is to consider approaching the point at which the derivative is presumed to exist along the real axis, and then similarly along the imaginary axis. The limit of the difference quotient has to be the same regardless of which direction we approach the point, and this gives us the necessary conditions. In general complex differentiation (of a complex variable) is a lot stronger than real differentiability, and the reason the basic rules of differentiation carry over to the complex case is due to algebraic properties, not analytic properties.

 

[DrRocket]

If z is complex, the following rules are true, right?

\frac{d}{dz}z^n = nz^{n-1}

\frac{d}{dz}\ln{g(z)} = \frac{1}{g(z)} \frac{d}{dz}g(z)

\frac{d}{dz}e^{g(z)}=e^{g(z)}\frac{d}{dz}g(z)

These are of course the same rules as for real variables.

When do I need to be careful about taking derivatives with respect to complex variables?

Do all of the same rules apply as with real variables? Or do some rules fail?

The usual manipulative rules for complex-valued functions work just like they do in elementary calculus.

The reason is that they are basically the same thing, with only a slight twist.

One can view a function of a complex variable as simply a function defined on \mathbb R^2 and apply the theory of differential calculus in several variables. The slight twist comes about when the the range is also the complex numbers and you demand that the derivative be a linear function over the complex field and not just over the real numbers. That is where the Cauchy-Riemann equations come into play. But the basic manipulative properties of the derivative remain in force, just as they do for a function from \mathbb R^2 to \mathbb R^2.

The magic of complex analysis only comes into play in terms of properties like the fact that a function that is once differentiable in the complex sense is in fact analytic in the sense of admitting a local power series, and hence derivatives of all orders.

The down side, if one can call it that, is that one cannot have smooth partitions of unity in the complex case, unlike in the real case. This is due to the lack of "bump functions" since any analytic function with compact support inside an open set is identically zero on that set.