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.
Best Answer
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.