Complex differentiation Is differentiation in the complex plane the same as that in the reals? In particular do the normal differentiation rules apply in the complex case such that I can just treat a complex map as a real map? 
Thanks. 
 A: I'm a little confused about your question, but I can think of two possible interpretations and I will try to handle them both.
The first interpretation: if $f(z)$ is a complex differentiable function and we have a formula for $f'(x)$ where $x$ is a real variable, can we recover the complex derivative of $f$ by "plugging in $z$ into that formula?  The answer is a qualified "yes".  The best precise statement along these lines is that if $f$ is complex differentiable at $z_0$ then it admits a Taylor series approximation $f(z) = \sum_{n=0}^\infty a_n z^n$ in a neighborhood of $z_0$ which converges uniformly on compact sets.  Uniform convergence allows us to differentiate term by term, and thus a formula for the derivative is given by $f'(z) = \sum_{n=1}^\infty n a_n z^{n-1}$.  So if your formula for calculating the real derivative of $f$ can be expressed in terms of Taylor series then it will be fine to simply "plug in" $z$.  
Note that you have to assume at the outset that $f$ is complex differentiable: even though you can calculate the derivative of $e^{-1/x}$ near $x=0$ using Taylor series, it doesn't extend to a complex differentiable function.  Also note that in practice one calculates real derivatives using things like the product rule and chain rule, and the same tools work in the complex case.  So often you can calculate the derivative of a complex differentiable function without resorting to Taylor series using the same technique that you use in the real case.
Second interpretation: If $f: \mathbb{C} \to \mathbb{C}$ is complex differentiable, is it true that its derivative agrees with the real derivative of $f$ regarded as a function $\mathbb{R}^2 \to \mathbb{R}^2$?  Here the answer is a more straightforward "yes", and it follows from the intrinsic definition of the derivative as the best linear approximation of $f$ near a point (which applies to both complex and real differentiable functions).  Again, we must assume at the outset that $f$ is complex differentiable for this to be correct: by the Cauchy-Riemann equations a real differentiable function at $(x_0,y_0)$ corresponds to a complex differentiable function if and only if the derivative is conformal, i.e.
$$df(x_0,y_0) = \left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)$$
A: The answer is yes if by "normal differentiation rules" you mean the sum, product, quotient and chain rule. However, there are very important differences. Some examples:


*

*If a function $f\colon D^{\text{open}}\subset\mathbb{C}\to\mathbb{C}$ is differentiable in the complex sense, then it is infinitely differentiable, even analytic. This is far from the truth for functions $f\colon(a,b)\to\mathbb{R}$, which can be differentiable with discontinuous derivative.

*If a function $f\colon D^{\text{open}}\subset\mathbb{C}\to\mathbb{C}$ is differentiable in the complex sense and vanishes on a set with a limit point in $D$, then $f\equiv0$. This again does not happen with differentiable functions of one variable: let $f(x)=e^{-1/x}$ if $x>0$, $f(x)=0$ if $x\le0$.

*If $f\colon\mathbb{C}\to\mathbb{C}$ is differentiable in the complex sense and bounded, then $f$ is constant. This again fails for real valued functions of a real variable: consider for instance $f(x)=\sin x$.

A: Here is a "rule" that fails: if $f$ is differentiable, then $\overline f$ is differentiable, and $\overline{f'}=(\overline f) '$.
This rule is true if the variable is real (for complex valued functions). It is also true if the complex plane is treated as $\mathbb{R}^2$ and differentiation as real differentiation on $\mathbb{R}^2$. So you might say that failure of this rule is what distinguishes complex analysis from real analysis.
