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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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Assume that $$f:\quad z=x+iy\quad \mapsto\quad w=u+iv$$is complex differentiable at the point $z_0=x_0+iy_0$, and let $${\bf f}:\quad(x,y)\mapsto(u,v)$$ be the corresponding ${\mathbb R}^2$-map. I claim that ${\bf f}$ is ${\mathbb R}$-differentiable at ${\bf z}_0=(x_0,y_0)$.

Proof. Let $f'(z_0)=:c=a+ib\in{\mathbb C}$. Then $$f(z_0+Z)-f(z_0)=c\>Z+o\bigl(|Z|\bigr)\qquad(Z\to0\in{\mathbb C})\ .$$ Separating real and imaginary parts here gives $$\eqalign{ u(x_0+X,v_0+Y)-u(x_0,y_0)&=aX-bY+o\bigl(\sqrt{U^2+V^2}\bigr)\cr v(x_0+X,v_0+Y)-v(x_0,y_0)&=bX+aY+o\bigl(\sqrt{U^2+V^2}\bigr)\cr}\qquad\bigl((U,V)\to(0,0)\in{\mathbb R}^2\bigr)\ .$$ This can be written as $${\bf f}({\bf z}_0+{\bf Z})-{\bf f}({\bf z}_0)=\left[\matrix{a&-b\cr b&a\cr}\right]\left[\matrix{X\cr Y\cr}\right]+o\bigl(|{\bf Z}|\bigr)\qquad({\bf Z}\to{\bf 0})$$ and proves $$\bigl[d{\bf f}({\bf z}_0)\bigr]=\left[\matrix{a&-b\cr b&a\cr}\right]\ .\qquad\square$$

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Is possible that the function is not differentiable across the real line, but is differentiable across any other direction in the complex plane.

There is no "real plane", so I take it as "real line".

Take the function $|x|$. It is not differentiable at $x=0$, but if you generate a surface by translating the function horizontally across the complex plane, it is constant across that direction, so it is differentiable in that direction, even on the tip of the function.

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You can smooth it fast enough on any direction (except the real line), to make it differentiable on any direction, except on the real line.

For example, the triangle function can be approximated by a fourier series. The constant line would be a single term in the fourier series, and the real line would have all the terms in the series. If you generate the surface by increasing the number of terms with the angle, from 0 terms at 90° to $\infty$ terms at 0°, the resulting surface would be differentiable on any direction, except the real line.