Welcome to our community

Be a part of something great, join today!

Proof of Argument Theorem

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
[tex]\text{I will prove the Argument THM }[/tex]

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

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

[tex]\text{Assume : }f(z) = \frac{P(z)}{Q(z)}[/tex]
[tex]f'(z) = \frac{P'(z)Q(z) - P(z) Q'(z)}{(Q(z))^2}[/tex]

[tex]\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 [/tex]

[tex]\frac{1}{2i\pi } \oint_{\gamma}\,\frac{P'(z)Q(z) - P(z) Q'(z)}{Q(z)P(z)}\, dz [/tex]

[tex]\frac{1}{2i\pi } \oint_{\gamma}\,\frac{P'(z)}{P(z)}-\frac{ Q'(z)}{Q(z)}\, dz [/tex]

[tex]\frac{1}{2i\pi } \oint_{\gamma}\,\frac{P'(z)}{P(z)}\,dz \, -\, \frac{1}{2i\pi}\oint_{\gamma}\,\frac{ Q'(z)}{Q(z)}\, dz [/tex]

[tex]\frac{1}{2i\pi } \oint_{\gamma}\,\frac{P'(z)}{P(z)}\,dz [/tex]

[tex]\text{Assume : }P(z) = (z-z_0)(z-z_1)...(z-z_N)[/tex]

[tex]P'(z) = (z-z_1)...(z-z_N)+(z-z_0)...(z-z_N)+...+(z-z_0)...(z-z_{N-1})[/tex]


[tex]\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)}[/tex]

[tex]\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 [/tex]

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

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


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

Feel free to post any comments .