7
$\begingroup$

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.

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

2 Answers 2

5
$\begingroup$

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, $$ \bbox[5px,border:2px solid black] { \frac{\partial F}{\partial y}(z) = i f(z). } $$

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – underfilho
    Commented Jun 18 at 14:26
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .