# Ahlfors' proof of Cauchy's theorem in a disk

I'm stuck in two parts of Ahlfors' proof of Cauchy's theorem in a disk (page 113), that is, if $f$ is holomorphic in an open disc $D$ then $\int_\gamma f(z)dz=0$ over every closed curve $\gamma$ in $D$.

First part:

Fix $z_0\in D$. We define $F(z)=\int_\sigma f(\zeta)d\zeta$ where $\sigma$ is the path joining $z_0$ with $z$ by taking an horizontal line from $z_0$ and getting to $z$ with a vertical line (hope it is clear).

"It is immediately seen that $\frac{\partial F}{\partial y}(z)=if(z)$". Not for me. I mean, I'm geometrically and intuitively inclined to understand that when we derive vertically, since it is a constant path what we should get is the value of $f$. I don't see how the $i$ appears, though, and I formally don't understand what's going on.

First of all, what does $\frac{\partial F}{\partial y}$ mean? This is what I understand by that: if $F=u+iv$ then $\frac{\partial F}{\partial y}=\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}$. Right?

I also found the following formula scribbled on my notebook: $\int_\gamma f(z)dz = \int_\gamma f(z) dx + i \int_\gamma f(z) dy$. Is this correct? I don't see how it makes sense to integrate with respect to $x$ a complex-valued function: what am I supposed to do with the $y$'s in the integrand? I'm guessing some abuse of notation is going on here.

Anyway, I'm guessing this is something easy and I'm just confused by notation.

Second part: we get that $\frac{\partial F}{\partial y}(z)=if(z)$ and $\frac{\partial F}{\partial x}(z)=f(z)$. "Hence $F$ is holomorphic with derivative $f$". How is that? I mean, we have $\frac{\partial F}{\partial x}= -i \frac{\partial F}{\partial y}$, which does not seem like Cauchy-Riemann to me. I'm guessing this is the same confusion as above.

Hope I was not overly verbose.

• In the imaginary direction $dz=idy$ so $\frac{\partial}{\partial z}=\frac{1}{i}\frac{\partial}{\partial y}$, hence $\frac{\partial F}{\partial y}=i \frac{\partial F}{\partial z} = i f(z)$.
– ulvi
Commented Jan 27, 2011 at 2:42
• @ulvi: Again, I'm inclined to believe that, as well that in the real direction, $dz=dx$, but I don't formally understand it. Commented Jan 27, 2011 at 18:44

### Why is $$\frac{\partial F}{\partial x} = -i \frac{\partial F}{\partial y}$$ called the Cauchy–Riemann equation for $$F(z)$$?

A function of the complex variable $$z = x + iy$$ is also sometimes considered as a function of the pair of real variables $$(x, y)$$. Thus, if $$F(z) = u(x, y) + i v(x, y)$$, then $$\frac{\partial F}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}\quad \text{and} \quad \frac{\partial F}{\partial y} = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}.$$ Now, the Cauchy–Riemann equations for $$F(z)$$ are $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$ On the other hand, $$\frac{\partial F}{\partial x} = -i \frac{\partial F}{\partial y}\ \iff \ \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} -i\frac{\partial u}{\partial y}.$$ Comparing the real and imaginary parts, we see that the two Cauchy–Riemann equations for $$F(z)$$ can also be concisely stated as $$\bbox[5px,border:2px solid black] { \frac{\partial F}{\partial x} = -i \frac{\partial F}{\partial y}. }$$

### Does it make sense to write $$\int_\gamma f(z)\, dz = \int_\gamma f(z)\, dx + i \int_\gamma f(z)\, dy$$?

Ahlfors defines the complex line integral of the continuous function $$f(z)$$ over a piecewise differentiable arc $$\gamma$$ as $$\int_\gamma f(z)\, dz = \int_a^b f(z(t)) z'(t)\, dt\label{defn}\tag{1}$$ where $$z = z(t)$$, $$a \leq t \leq b$$, is a parametrization of the arc $$\gamma$$. This is given in $$\S$$4.1.1 on page 102.

On the next page, Ahlfors defines line integrals with respect to $$\bar{z}$$ as $$\int_\gamma f\, \overline{dz} = \overline{\int_\gamma \bar{f}\, dz}.$$ Finally, line integrals with respect to $$x$$ and $$y$$ are defined as \begin{align} \int_\gamma f\, dx &= \frac{1}{2} \left( \int_\gamma f\, dz + \int_\gamma f\, \overline{dz} \right), \\ \int_\gamma f\, dy &= \frac{1}{2i} \left( \int_\gamma f\, dz - \int_\gamma f\, \overline{dz} \right). \end{align} Then, one sees that we indeed have $$\bbox[5px,border:2px solid black] { \int_\gamma f\, dz = \int_\gamma f\, dx + i \int_\gamma f\, dy. }\label{zxy}\tag{2}$$ Furthermore, using \eqref{defn} one can show the analogous formulas \begin{align} \int_\gamma f\, dx &= \int_a^b f(z(t)) x'(t)\, dt, \label{xt}\tag{3} \\ \int_\gamma f\, dy &= \int_a^b f(z(t)) y'(t)\, dt, \label{yt}\tag{4} \end{align} where $$z(t) = (x(t),y(t))$$, $$a \leq t \leq b$$, is a parametrization of the arc $$\gamma$$.

### How is it immediate that $$\partial F / \partial y = i f(z)$$?

Here, $$F(z)$$ is defined as $$F(z) = \int_\sigma f\, dz$$ where $$\sigma$$ is the horizontal line segment from $$(x_0, y_0)$$ to $$(x, y_0)$$ followed by the vertical line segment from $$(x, y_0)$$ to $$(x, y)$$. Now, write $$F(z)$$ as $$F(z) = \int_{\sigma_1} f\, dz + \int_{\sigma_2} f\, dz,$$ where $$\sigma_1$$ is the horizontal line segment from $$(x_0, y_0)$$ to $$(x, y_0)$$, and $$\sigma_2$$ is the vertical line segment from $$(x,y_0)$$ to $$(x,y)$$. Then, further expanding each integral on the right using \eqref{zxy}, we get $$F(z) = \int_{\sigma_1} f\, dx + i\int_{\sigma_2} f\, dy,$$ since $$\int_{\sigma_1} f\, dy = 0 = \int_{\sigma_2} f\, dx$$ (use \eqref{xt} and \eqref{yt} to convince yourself that this is indeed so).

So, consider the values $$F(z(k))$$ where $$z(k) = x + i(y + k)$$. Note that $$z(0) = z$$. By the definition of partial derivative, we have $$\frac{\partial F}{\partial y}(z) = \lim_{k \to 0} \frac{F(z(k)) - F(z)}{k} = i \lim_{k \to 0} \frac{1}{k} \int_{z}^{z(k)} f\, dy,$$ where the last integral is along the straight line segment from $$z = (x,y)$$ to $$z(k) = (x, y+k)$$. Using \eqref{yt}, we have $$\int_{z}^{z(k)} f\, dy = \int_{y}^{y+k} f(x,t) \cdot 1\, dt,$$ so by the first fundamental theorem of calculus, the limit is precisely $$f(z)$$. Thus, $$\bbox[5px,border:2px solid black] { \frac{\partial F}{\partial y}(z) = i f(z). }$$

• why though that same method does not work for $\frac{\partial F}{\partial x}$ in that same curve, and we, instead, need to replace by the other sides of the rectangle? I don't get that, for me it's like we can do the same using only these two sides of the rectangle and we didn't needed cauchy theorem for rectangles. Commented Jun 18 at 14:26

Second part: The Cauchy-Riemann equations are equivalent to $\frac{\partial F}{\partial y}(z) = if(z)$. This form dictates when a function is conformal. In particular, look at the matrix $$\begin{bmatrix} u_x & -v_x \\ v_x & \ \ \ u_x \end{bmatrix}$$

This is precisely the matrix representation of a complex number.