Proof of Argument Theorem

ZaidAlyafey

Well-known member
MHB Math Helper
$$\text{I will prove the Argument THM }$$

$$\text{If f(z) is a function that is analytic in a closed simple curved }\gamma$$
$$\text{ and has M Poles , and N Zeors inside }\gamma$$

$$\text{Then : }\frac{1}{2i\pi } \oint_{\gamma}\, \frac{f'(z)}{f(z)}\, dz \,=\, N-M$$

$$\text{Assume : }f(z) = \frac{P(z)}{Q(z)}$$
$$f'(z) = \frac{P'(z)Q(z) - P(z) Q'(z)}{(Q(z))^2}$$

$$\text{R.H }\frac{1}{2i\pi } \oint_{\gamma}\, \frac{f\frac{P'(z)Q(z) - P(z) Q'(z)}{(Q(z))^2}}{\frac{P(z)}{Q(z)}}\, dz$$

$$\frac{1}{2i\pi } \oint_{\gamma}\,\frac{P'(z)Q(z) - P(z) Q'(z)}{Q(z)P(z)}\, dz$$

$$\frac{1}{2i\pi } \oint_{\gamma}\,\frac{P'(z)}{P(z)}-\frac{ Q'(z)}{Q(z)}\, dz$$

$$\frac{1}{2i\pi } \oint_{\gamma}\,\frac{P'(z)}{P(z)}\,dz \, -\, \frac{1}{2i\pi}\oint_{\gamma}\,\frac{ Q'(z)}{Q(z)}\, dz$$

$$\frac{1}{2i\pi } \oint_{\gamma}\,\frac{P'(z)}{P(z)}\,dz$$

$$\text{Assume : }P(z) = (z-z_0)(z-z_1)...(z-z_N)$$

$$P'(z) = (z-z_1)...(z-z_N)+(z-z_0)...(z-z_N)+...+(z-z_0)...(z-z_{N-1})$$

$$\frac{P'(z)}{P(z)} =\frac{ (z-z_1)...(z-z_N)+(z-z_0)...(z-z_N)+...+(z-z_0)...(z-z_{N-1})}{(z-z_0)(z-z_1)...(z-z_N)}$$

$$\frac{1}{2i\pi } \oint_{\gamma}\, \frac{ (z-z_1)...(z-z_N)+(z-z_0)...(z-z_N)+...+(z-z_0)...(z-z_{N-1})}{(z-z_0)(z-z_1)...(z-z_N)}\,dz$$

$$\frac{1}{2i\pi } \oint_{\gamma}\,\frac{1}{z-z_0}+\frac{1}{z-z_1}+...+\frac{1}{z-z_N} = N$$

$$\text{similarily we can show : }\frac{1}{2i\pi}\oint_{\gamma}\,\frac{ Q'(z)}{Q(z)}\, dz = M$$

$$\text{Then : }\frac{1}{2i\pi } \oint_{\gamma}\, \frac{f'(z)}{f(z)}\, dz \,=\, \fbox{N-M} \text{ W.R.T}$$

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