Also sprach Zarathustra said:Explain me please how to show that C.R conditions are holds.
$$f(z)=\left\{\begin{matrix}e^{-\frac{1}{z^4}}\ \ \ \text{if z not 0 }\\ 0 \ \ \ \ \ \ \ \text{if z not 0 }\end{matrix}\right.$$Thank you.
I think that the point of this question is to see what happens at the origin (see here).Also sprach Zarathustra said:Explain me please how to show that C.R conditions are holds.
$$f(z)=\begin{cases}e^{-1/z^4}&(z \ne 0)\\ 0 & (z = 0) \end{cases}$$
You are right, about the purpose of this question!I still have troubles make $f(z)$ into the form $f(x,y)=u(x,y)+v(x,y)$.Opalg said:I think that the point of this question is to see what happens at the origin (see here).
For every nonzero value of $z$, the function $f(z)$ is analytic in a neighbourhood of $z$ and therefore satisfies the C–R equations. The surprising thing is that $f(z)$ also satisfies the C–R equations at $z=0$, despite not being analytic (not even continuous) there.
To see that, write $z=x+iy$ and $f(z) = f(x,y) = u(x,y) + iv(x,y)$. Then $$\left.\frac{\partial u}{\partial x}\right|_{(0,0)} = \lim_{x\to0}\frac{(\text{re }f(x,0)) - (\text{re }f(0,0))}{x} = \lim_{x\to0}\frac{\text{re }e^{-1/x^4}-0}x.$$
That limit is 0. Similar calculations show that $\frac{\partial v}{\partial x}$, $\frac{\partial u}{\partial y}$ and $\frac{\partial v}{\partial y}$ are all 0 at the origin, and therefore the C–R equations are satisfied there.
In fact, you do not need to find explicit formulas for $u(x,y)$ and $v(x,y)$. To evaluate $\displaystyle\frac{\partial u}{\partial x}$ at the origin, you have to go back to the definition of the derivative $$\left.\frac{\partial u}{\partial x}\right|_{(0,0)} = \lim_{x\to0}\frac{(\text{re }f(x,0)) - (\text{re }f(0,0))}{x}$$ (as in my previous comment). That limit is equal to $$\lim_{x\to0}\frac{\text{re }e^{-1/x^4}-0}x.$$ But $e^{-1/x^4}$ is already real, so its real part is itself. Therefore $$\left.\frac{\partial u}{\partial x}\right|_{(0,0)} = \lim_{x\to0}\frac{e^{-1/x^4}}x.$$ That is a purely real limit, which is equal to 0 (essentially because exponential convergence is more powerful than polynomial convergence).Also sprach Zarathustra said:I still have troubles make $f(z)$ into the form $f(x,y)=u(x,y)+v(x,y)$.
$$f(x,y)=e^{x^4-6x^2y^2-y^4}e^{4i(xy^3-x^3y)}$$
C.R conditions refer to the conditions of continuity and differentiability that must be satisfied in order to apply the Cauchy-Riemann equations in complex analysis.
To show that C.R conditions are held, we must first check that the partial derivatives of the real and imaginary parts of a complex function exist and are continuous at a given point. We also need to verify that these partial derivatives satisfy the Cauchy-Riemann equations at that point.
C.R conditions are important because they are necessary for a complex function to be differentiable at a given point. If these conditions are not satisfied, the Cauchy-Riemann equations cannot be used and the function is not considered differentiable at that point.
If C.R conditions are not held, then the complex function is not differentiable at that point. This means that the function does not have a well-defined derivative and cannot be approximated by a linear function at that point.
Yes, it is possible for C.R conditions to be held at some points and not at others. This depends on the specific function and its properties at each point. It is important to check C.R conditions at each point of interest in order to determine the differentiability of the function.