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.

I asked the question in the title.

Is it possible for a function to be differentiable in the complex plane but not in the real plane?

Could you help me find some examples or explain how it is possible?

share|improve this question
2  
One way to see how this is possible is by interpreting the derivative as a linear transformation. In the complex plane, the derivative is essentially a rotation matrix (thanks to the CR equations), which is certainly not the case in $\mathbb{R}^2$. –  John Martin Oct 29 '12 at 2:24
5  
Correct me if I'm wrong, but if a function $f:\mathbb{C} \to \mathbb{C}$ is differentiable, then its component functions are both $C^{1}$ (in fact they are $C^{\infty}$). It follows that $f$ is differentiable when viewed as a function from $\mathbb{R}^2 \to \mathbb{R}^2$. –  littleO Oct 29 '12 at 2:38
    
@JohnMartin I am going to need to be told a bit more details, but I think I get the main topic. –  yiyi Oct 29 '12 at 2:40
2  
So, what littleO is saying is that if you have a complex function $f$ of complex variable, then we can write $f(x,y) = u(x,y) + iv(x,y)$, where $u, v$ are the component functions. Now, if $f$ is differentiable, then the partials of $u, v$ w.r.t $x, y$ exist and are continuous. Note that these partials also satisfy the Cauchy-Riemann equations. But, then continuity of partials implies the total derivative of $f$ exists, if $f$ is viewed as a function from an open set in $\mathbb{R}^2$ to $\mathbb{R}^2$. So, $f$ is differentiable as a function of $\mathbb{R}^2$. –  Rankeya Oct 29 '12 at 2:48
2  
Dear @MaoYiyi: Yes, it not possible. Complex differentiability is a very strong condition. For instance, if a function is holomorphic, then all its derivatives are also holomorphic, i.e., it is infinitely differentiable. –  Rankeya Oct 29 '12 at 3:21

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.