## winding number questions

THEOREM
If the piecewise differentiable closed curve $$\gamma$$ does not pass through the point a then:

$$(1) \displaystyle\oint_{\gamma}\frac{dz}{z-a}$$

is s multiple of $$2 \pi i$$

PROOF
$$\gamma$$ is given by z(t) $$\alpha \leq t \leq \beta$$ consider:

$$(2) h(t)=\displaystyle\int^{t}_{\aplha}\frac{z'(t)}{z(t)-a}dt$$

h(t) is defined and continuos on the closed interval $$(\alpha, \beta)$$

$$(3) h'(t)=\frac{z'(t)}{z(t)-a}$$

where z'(t) is continuos.

$$(4) k=e^{-h(t)}(z(t)-a)$$

(5) Hence k' = 0.

$$(6) e^{h(t)}=\frac{(z(t)-a)}{k}$$

$$(7) e^{h(t)}=\frac{(z(t)-a)}{z(\alpha)-a}$$

Since $$\gamma$$ is a closed curve $$z(\beta)=z(\alpha)$$

$$(8) e^{h(\beta)}=1$$

$$h(\beta)$$ must be a multiple of $$2 \pi i$$.

END

QUESTIONS

1. (3) why do we need to know about h'(t)?

2. (5) why is k' = 0?

3. How did we replace k with $$z(\alpha)-a$$ in step (7) ?

Thanks for any help you can give. I need to understand this proof for a test and I'd rather not memorize these bits, but instead know what I'm doing! Thanks!
 You need to know $h^\prime (t)$ in order to answer question 2. For question 2, calculate $\frac{dk}{dt}$. You should get $$\frac{dk}{dt} = -h^\prime (t) e^{-h(t) } (z(t) -a ) + e^{-h (t)} z^\prime (t)$$ And substitute what you're given for $h^\prime (t)$.
 Thanks! I think I get it now!

## winding number questions

I'm not 100% about part 3, but I think it's because $\frac{dk}{dt}=0$. Thus we know that k must be a constant. Since k is a constant, if we know it's value at one place, we know it at any other. Thus $k \equiv k(\alpha) = e^{-h(\alpha)} ( z(\alpha) - a)$ but then there's an extra $e^{-h(\alpha)}$ and I'm not sure where it disappears.