- 269

- 0

THEOREM

If the piecewise differentiable closed curve [tex]\gamma[/tex] does not pass through the point

**a**then:

[tex](1) \displaystyle\oint_{\gamma}\frac{dz}{z-a}[/tex]

is s multiple of [tex]2 \pi i[/tex]

PROOF

[tex]\gamma[/tex] is given by z(t) [tex]\alpha \leq t \leq \beta[/tex] consider:

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

h(t) is defined and continuos on the closed interval [tex](\alpha, \beta)[/tex]

[tex](3) h'(t)=\frac{z'(t)}{z(t)-a}[/tex]

where z'(t) is continuos.

[tex](4) k=e^{-h(t)}(z(t)-a)[/tex]

(5) Hence k' = 0.

[tex](6) e^{h(t)}=\frac{(z(t)-a)}{k}[/tex]

[tex](7) e^{h(t)}=\frac{(z(t)-a)}{z(\alpha)-a}[/tex]

Since [tex]\gamma[/tex] is a closed curve [tex]z(\beta)=z(\alpha)[/tex]

[tex](8) e^{h(\beta)}=1[/tex]

[tex]h(\beta)[/tex] must be a multiple of [tex]2 \pi i[/tex].

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 [tex]z(\alpha)-a[/tex] 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!