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Are the differentiability in the complex plane and differentiability in $\Bbb R^2$ different concepts?

Consider the linear operator $T$ on $\Bbb R^2$ defined by $T(x,y) = (x+y,x-y),\ x,y \in \Bbb R$.Then clearly $T$ is differentiable in $\Bbb R^2$ but it is not differentiable in $\Bbb C$ since Cauchy-Riemann equation is not satisfied at any point of $\Bbb C$.

What is the basic difference between these two notions of differentiability?

Any help will be highly appreciated.

little o
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2 Answers2

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This is all about linear transformations from $\mathbb {R} ^2$ to $\mathbb{R^2}$. A typical transformation $T$ is given by a matrix $$M_{T}=\begin{bmatrix}a & b\\c & d\end{bmatrix} $$ such that $$T(x, y) =M_{T} \begin{bmatrix} x\\y\end{bmatrix} $$ or $$T(x, y) =(ax+by, cx+dy) $$ where $a, b, c, d$ are real.

On the other hand a linear map from $\mathbb{C} $ to $\mathbb{C} $ is always given by $f(z) =cz$ where $c=a+ib$ is some complex number. If $z=(x+iy) $ then this means $$f(z) =ax-by+i(bx+ay) $$ If we try to represent this as a linear transformation from $\mathbb{R}^2$ to itself then the matrix of this transformation is $$M_{f} =\begin{bmatrix} a & -b\\b & a\end{bmatrix} $$ and further this means that any linear transformation $M_{T} $ given earlier acts as a complex linear transformation if and only if $a=d, b=-c$.

The difference between real differentiability and complex differentiability of $$f(z) =u(x, y) +iv(x, y) $$ is all about knowing when the usual derivative transformation (Jacobian) $$D_{f} =\begin{bmatrix} \dfrac{\partial u} {\partial x} & \dfrac{\partial u} {\partial y} \\ \dfrac{\partial v} {\partial x} & \dfrac{\partial v} {\partial y} \end{bmatrix} $$ can be viewed as a complex linear transformation and then one immediately gets the famous Cauchy Riemann equations.

Paramanand Singh
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  • So in conclusion we may say that the complex differentiability is stronger than differentiability in $\Bbb R^2$ since every complex linear transformation is a linear transformation on $\Bbb R^2$ but not conversely. Am I right @Paramanand Singh? – little o Nov 09 '18 at 04:41
  • @Dbchatto67: yes you are correct. – Paramanand Singh Nov 09 '18 at 05:33
  • Thank you very much @Paramanand Singh for your help. Very nice answer. – little o Nov 09 '18 at 05:41
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One can prove the following theorem (see any basic complex analysis textbook):

The following 2 statements are equivalent for a function $$f:A \subseteq \mathbb{C} \to \mathbb{C}$$ given by

$$f(z) = f(x+yi) = u(x,y) + iv(x,y)$$

(1) $ f $ is complex differentiable in $a = c + di \in A$.

(2) $ u,v: V \subseteq \mathbb{R}^2 \to \mathbb{R}$ are differentiable (in the multivariate sense) in $(c,d)$ AND $f$ satisfies the Cauchy-Riemann equations in $a$.

From this , one sees that complex differentiability is much stronger than regular differentiability.

  • Look at this link https://math.stackexchange.com/a/1640257/543867 – little o Nov 08 '18 at 07:24
  • Yes? What's with the link? –  Nov 08 '18 at 07:25
  • This link gives an opposite interpretation of what you have answered. Look at the Christian Blatter's answer. – little o Nov 08 '18 at 07:27
  • You are confusing things. It is a special case indeed, one where an EXTRA condition is imposed, which are the Cauchy Riemann equations (this is the special matrix in his answer). So, to be complex differentiable you must be regularly differentiable and the matrix has to have a special form, which is strictly stronger than only requiring regular differentiability. –  Nov 08 '18 at 07:29
  • Does complex differentiability imply $\Bbb R^2$-differentiability? – little o Nov 08 '18 at 07:39
  • Yes. My answer says it implies differentiability of the component functions, so if you view $f$ as a function $\mathbb{R}^2 \to \mathbb{R}^2$ it is differentiable too. –  Nov 08 '18 at 07:47