Complex analysis derivatives

Question

If a function $f(x,y)=u(x,y)+i\cdot v(x,y)$ is $\mathbb{C}$-differentiable,how can I prove that for $u$ and $v$ the partial derivatives exist? I have figured out if f is continuously differentiable then the derivative value must be $f_x$,but $f_x$ exists only when $u_x,v_x$ exist.So how can I ensure their existence?

Answer 1

Let $z_0\in\mathbb C$. Asserting that $f$ is differentiable at $z_0$ means that the lime$$\lim_{h\to0}\frac{f(z_0+h)-f(z_0)}h$$ exists. In particular, the limit$$\lim_{h\to0,\ h\in\mathbb{R}}\frac{u(z_0+h)+iv(z_0+h)-u(z_0)-if(z_0)}h$$exists, which is the same thing as asserting that both limits$$\lim_{h\to0,\ h\in\mathbb{R}}\frac{u(z_0+h)-u(z_0)}h\text{ and }\lim_{h\to0,\ h\in\mathbb{R}}\frac{v(z_0+h)-v(z_0)}h$$exist. But this means that $u_x$ and $v_x$ exist at $z_0$.

A similar argument shows that $u_y$ and $v_y$ exist at $z_0$.

Answer 2

Unpack the definition; the function is $\mathbb{C}$-differentiable if $\lim_{z\to z_0} \frac{f(z)-f(z_0)}{z-z_0} = f'(z_0)$ exists. $$\lim_{z\to z_0} \frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) = 0$$ $$\lim_{z\to z_0} \frac1{z-z_0}\left(f(z)-f(z_0) - f'(z_0)(z-z_0)\right) = 0$$ $$\lim_{z\to z_0} \frac1{|z-z_0|}\left|f(z)-f(z_0) - f'(z_0)(z-z_0)\right| = 0$$ $$\lim_{(x,y)\to (x_0,y_0)}\frac1{\|(x,y)-(x_0,y_0)\|}\left\|u(x,y)+iv(x,y)-u(x_0,y_0)-iv(x_0,y_0) - L_{f'(z_0)}(x-x_0,y-y_0)\right\| = 0$$ There $L_{\zeta}$ is the linear map from $\mathbb{R}^2$ to $\mathbb{C}$ that acts the same as multiplication by $\zeta$. Split this into real and imaginary parts; we get $$\lim_{(x,y)\to (x_0,y_0)}\frac1{\|(x,y)-(x_0,y_0)\|}\left\|u(x,y)-u(x_0,y_0)- L(x-x_0,y-y_0)\right\| = 0$$ for some linear transformation $L$, and similarly for the imaginary part. That's the definition of differentiability for a function from $\mathbb{R}^2$ to $\mathbb{R}$; we have proved, straight from the definition, that if a function is $\mathbb{C}$-differentiable, its real and imaginary parts are $\mathbb{R}$-differentiable. A $\mathbb{R}$-differentiable function has partial derivatives, and we're done.

Next, of course, is verifying the relation between those partial derivatives (the Cauchy-Riemann equations). Related, the theorem to remember from this: a function is $\mathbb{C}$-differentiable at a point if and only if it's $\mathbb{R}$-differentiable at that point and its partial derivatives there satisfy the Cauchy-Riemann equations.

Yes, we could have gotten just the partial derivatives more easily. The full power of differentiability at that point is worth a little extra effort.