Winding Number Proof Clarification

[futurebird]

I have some questions about this proof. I have numbered the equations (1), (2), ... so I can ask about them.


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!

 

[Kreizhn]

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).

 

[futurebird]

Thanks! I think I get it now!

 

[Kreizhn]

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.