We were asked if the function: $$ f(z) = \left\{ \begin{array}{ll} e^{-1/z^4} & \quad z \neq 0 \\ 0 & \quad z = 0 \end{array} \right. $$
satisfies the Cauchy-Riemann equations at $z = 0$. I don't know how to separate into the form $f(z) = u(x,y) + i v(x,y)$, and polar form gave me an undefined limit. The answer is apparently yes, but I'm not satisfied with the explanations I've seen thus far.
Before I tackle that problem, I thought I'd start with a simpler version:
$$ f(z) = \left\{ \begin{array}{ll} e^{-1/z} & \quad z \neq 0 \\ 0 & \quad z = 0 \end{array} \right. $$
Then, for $z \neq 0$,
$$ e^{\frac{-1}{z}} = e^{\frac{-1}{x+iy}} = e^{\frac{-1}{x+iy}\cdot \frac{x-iy}{x-iy}} = e^{\frac{-x}{x^2+y^2}\cdot} e^\frac{iy}{x^2+y^2} = e^{\frac{-x}{x^2+y^2}} \cdot \left( cos\left(\frac{y}{x^2+y^2} \right) + i \ sin \left( \frac{y}{x^2+y^2} \right) \right) $$
Then we can write $f(z) = u(x,y) + i v(x,y)$, where
$$ u(x,y) = \left\{ \begin{array}{ll} e^{\frac{-x}{x^2+y^2}} \cdot \left( cos\left(\frac{y}{x^2+y^2} \right)\right) & \quad (x,y) \neq (0,0) \\ 0 & \quad (x,y) = (0,0) \end{array} \right. $$ $$ v(x,y) = \left\{ \begin{array}{ll} e^{\frac{-x}{x^2+y^2}} \cdot \left( sin\left(\frac{y}{x^2+y^2} \right)\right) & \quad (x,y) \neq (0,0) \\ 0 & \quad (x,y) = (0,0) \end{array} \right. $$
Now, taking the partial derivatives at $z = 0$,
$$ u_x = \lim_{\Delta x\to 0} \frac{u(0 + \Delta x, 0) - u(0,0)}{\Delta x} = \lim_{\Delta x\to 0} \frac{e^\frac{-\Delta x}{(\Delta x)^2} cos(0) - 0}{\Delta x} = \lim_{\Delta x\to 0} \frac{e^\frac{-1}{\Delta x}}{\Delta x} = 0 $$ $$ v_y = \lim_{\Delta y\to 0} \frac{v(0, 0 + \Delta y) - v(0,0)}{\Delta y} = \lim_{\Delta y\to 0} \frac{e^0 sin(\frac{1}{\Delta y}) - 0}{\Delta y} = \lim_{\Delta y\to 0} \frac{sin(\frac{1}{\Delta y})}{\Delta y} $$
Did I take the partial derivatives correctly, or make a mistake elsewhere? Otherwise, I believe the limit for $v_y$ diverges (I tested in Matlab for small $\Delta y$ and does not converge). It makes intuitive sense since the numerator is bouncing between $-1$ and $1$ and the denominator keeps getting smaller, so I expect this limit does diverge. But that would mean the Cauchy-Riemann equations are not satisfied, and my intuition is telling me that is wrong.