Cauchy Theorem for a disk help to undersand

Question

I have problem to understand proof of the following theorem: If $f$ is analytic in an open disk $U$, then $\int_{\gamma}f(z)dz=0$ for every closed curve $\gamma$ in $U$.

Proof: We consider the disk $U$ centered in 0 and the point $z=x+iy$. Furthermore let $\gamma$ be the arc that is horizonzal form $(0,0)$ to $(x,0)$, and vertical from $(x,0)$ to $(x,y)$. We define now $$F(z)= \int_{\gamma}f(z)dz$$ We get $$\frac{\partial F}{\partial y}=if(z)$$ Now why is this true? I don't see how to proof this statement. Can someone pls helps me? Thanks!

Answer 1

$$\displaystyle \frac{\partial F}{\partial y}(z) = \lim_{h\rightarrow 0} \frac{F(x,y+h)-F(x,y)}{h} = \lim_{h\rightarrow 0} \frac{1}h \int_{\gamma^\star} f(w) \, dw$$ where $\gamma^\star$ is a vertical path from $(x,y)$ to $(x,y+h)$.

Since $f$ is continuous at $z$

$$f(w) = f(z) + \psi(w)$$ where $\psi(w) \rightarrow 0$ as $w\rightarrow z$

$$\int_{\gamma^\star} f(w) \, dw = \int_{\gamma^\star}f(z) \, dw+\int_{\gamma^\star} \psi(w)\, dw $$

The second integral goes to zero and $$\displaystyle \frac{\partial F}{\partial y}(z) = f(z) \lim_{h \rightarrow 0} \frac{1}{h} \int_{\gamma^\star} dw = if(z).$$

(Compare Stein & Shakarchi pp. 38, 39.)

Answer 2

No wonder you're confused. The proof is pretty sloppy.

Proof: We consider the disk $U$ centered in 0 and the point $z=x+iy$.

Let's change that to $z_0 = x_0 + i y_0$, because we're going to use the letter $z$ as an integration variable in a moment. And let's also say

... and the point $z_0 = x_0 + i y_0$, where $|z_0| \le 1$, so that $z_0$ is in the unit disk.

Furthermore let $\gamma$ be the arc that is horizontal from $(0,0)$ to $(x_0,0)$, and vertical from $(x_0,0)$ to $(x_0,y_0)$.

Let's add:

Note that $\gamma$ is an arc in $D$, because each of these segments is entirely contained in $D$, for $x_0 \le 1$ (why?), and $x_0, ty_0) \in D$ for $0 \le t \le 1$ (why?).

[The "why" in both cases, comes from the triangle inequality, or just some algebra playing around with $x_0^2 + y_0^2 \le 1$]

The curve $\gamma$ (presumably a function $\gamma:[0, 1] \to D$) actually depends not only on its argument, but also on $z_0$. With some reluctance, I'd rewrite this as:

Furthermore let $\gamma_{z_0}$ be the arc that is horizontal from $(0,0)$ to $(x_0,0)$, and vertical from $(x_0,0)$ to $(x_0,y_0)$. Note that $\gamma_{z_0}$ is an arc in $D$, because each of these segments is entirely contained in $D$, for $x_0 \le 1$ (why?), and $x_0, ty_0) \in D$ for $0 \le t \le 1$ (why?).

The author now writes

We define now $$F(z)= \int_{\gamma}f(z)dz$$

...and at this point, things are utterly meaningless, for $z$ now means two, or possibly three, different things. In our rewritten version, we have

We define now $$F(z_0)= \int_{\gamma_{z_0}}~f(z)~dz$$

where $z$ is the variable of integration, and $z_0$ is an arbitrary point of the disk. The function $F$ is, of course, a function of a single complex variable, but the author treats it with the usual "pun" as being a function of two real variables. Far better would be to write

Let $H : \{(x, y) \mid x^2 + y^2 \le 1 \} \to \Bbb C$ be defined by $$ H(x, y) = F(x + iy). $$

We get $$\frac{\partial F}{\partial y}=if(z)$$

In our version, this becomes:

We get $$\frac{\partial H}{\partial y}(x_0, y_0)=if(x_0 + iy_0).$$

or (perhaps better)

We get $$\frac{\partial H}{\partial y}(a, b)=if(a + ib).$$.=

You, quite reasonably, ask:

Now why is this true? I don't see how to proof this statement. Can someone pls helps me? Thanks!

Well (following @mjw's answer), \begin{align} \frac{\partial H}{\partial y}(a, b) & = \lim_{h \to 0} \frac{1}{h}(H(a, b+h) - H(a,b)) \end{align} The two $H$ expressions represent integrals over two very similar paths, which you can see by drawing them. Both go horizontally from $(0,0)$ to $(a, 0)$; both go vertically from $(a, 0)$ to $(a, b)$; the first one goes on all the way to $(a, b+h)$. The difference is therefore the integral of $f$ along a path from $a + bi$ to $a + (b+h)i)$; such a path is $$ \alpha(t) = a + (b + ht)i ~~~ 0 \le t \le 1 $$ So we have \begin{align} \frac{\partial H}{\partial y}(a, b) & = \lim_{h \to 0} \frac{1}{h}(H(a, b+h) - H(a,b))\\ & = \lim_{h \to 0} \frac{1}{h}(\int_{\alpha} f(z) ~ dz)\\ & = \lim_{h \to 0} \frac{1}{h}(\int_0^1 f(a + (b + ht)) \alpha'(t) ~ dt)\\ & = \lim_{h \to 0} \frac{1}{h}(\int_0^1 f(a + (b + ht)) ih ~ dt)\\ & = \lim_{h \to 0} i(\int_0^1 f(a + (b + ht)) ~ dt) \end{align} Now $$ f(a + (b + ht)i) \approx f(a + bi) + f'(a + bi)hi $$ using the Taylor series, with the approximation getting better and better as $h \to 0$. [NB: there's a slightly more rigorous argument using the mean value theorem, but I'm not going to write it out here.]

So continuing, we get \begin{align} \frac{\partial H}{\partial y}(a, b) & = \lim_{h \to 0} i(\int_0^1 f(a + (b + ht)) ~ dt)\\ & \approx \lim_{h \to 0} i(\int_0^1 f(a+ bi) + f'(a + bi)hi ~ dt)\\ & = \lim_{h \to 0} i(\int_0^1 f(a+ bi) dt + \int_0^1 f'(a + bi)hi ~ dt)\\ & = \lim_{h \to 0} i(f(a+ bi) + h\int_0^1 if'(a + bi) ~ dt)\\ & = \lim_{h \to 0} i(f(a+ bi) + ihf'(a + bi) )\\ & = if(a+ bi) + \lim_{h \to 0} i^2hf'(a + bi) \\ & = if(a+ bi) - f'(a + bi)\lim_{h \to 0} h \\ & = i~f(a+ bi). \end{align}

I may have lost an "i" or two in there, but that's the gist.

You might want to find a different intro-to-complex-variables to read. Sigh.