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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?

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    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 $\mathbb{R}^2$.2012-10-29
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    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 $\mathbb{R}^2 \to \mathbb{R}^2$.2012-10-29
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    @JohnMartin I am going to need to be told a bit more details, but I think I get the main topic.2012-10-29
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    @littleO Could you write more explaining what you mean.2012-10-29
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    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 $\mathbb{R}^2$.2012-10-29
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    @Rankeya so the answer to my question is no.2012-10-29
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    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.2012-10-29
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    @Rankeya Wow, going to have to think about that a bit more. I know that the Complex numbers are complete (no gaps), but I know that the Real numbers have gaps. So, if the function is differentiable in the Complex, then its differentiable in all the subsets numbers; but what if it is only continous in the complex, does that imply that its continous in the subsets number systems?2012-10-29
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    @littleO you are exactly right. My error. I was saying that real analytic functions are not necessarily complex analytic but this is pretty obvious, and also not the what is being asked.2012-10-29
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    @JohnMartin At first I had no idea of exactly what you were talking about, but after searching for the terms, I believe I now understand.2012-10-29
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    Dear @MaoYiyi: If you believe you have understood what littleO was trying to say, why don't you post an answer. I think it is a good way to test one's understanding.2012-10-30
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    @Rankey There is a huge vast chasm between, I think I know and I feel sure in telling others what I know. (I am not very clear upon the $C^1$ and $C^{\infty}$) I get the idea that if its differentiable in C then its differentiable in R2012-10-30
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    @Rankeya Do you have a book that could explain the difference between $C^1$ and $C^{\infty}$?2012-11-01

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