Cauchy's Formula

Question

In the proof for Cauchy's Formula in my notes it states that $$\frac{1}{2\pi i}\int_{∂D}\frac{dw}{w-z}=1,$$ can somebody please explain why this is to me. Is it something to do with Euler's identity?

If anybody could help me it would be much appreciated, sorry if this is not explained very well.

Thanks very much,

Answer 1

You can suppose that $\partial D$ is the circle of center $z$ and radius $r$, parametrize it $w = z + re^{it}$, $t\in[0,2\pi]$ and calculate directly the integral: $$\frac{1}{2\pi i}\int_{|w - z| = r}\frac{dw}{w-z} = \cdots$$ Can you continue?

Answer 2

$$ \int_{\partial D} \frac {dw} w $$

For the moment let's suppose $\partial D$ is the circle of unit radius centered at $0$ and you're going around in the counterclockwise direction. Think of $dw$ as an infinitely small change in $w.$ The direction in which you move around the circle is alwasy $90^\circ$ counterclockwise from the direction from $0$ out to $w.$ Multiplication by $i$ is a $90^\circ$ counterclockwise rotation. Therefore $\dfrac{dw} w = i \, \left| \dfrac{dw} w\right| = i\,\left|dw\right|.$ Hence $$ \int_{\partial D} \frac{dw} w = \int_{\partial D} i\,\left|dw\right|. $$ The value of the integral is then $i$ times the length of the curve.

Now supposing $D$ is not the unit disk centered at $0.$ There's a short intuitive argument, which can be made rigorous, showing that integrals along curves homotopic to each other are equal. The reason the boundary of the disk is not homotopic to zero is that it winds around the origin, which is a point that is not in the domain.