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Suppose ff is differentiable at points on a closed path γ\gamma and at all points in the region G enclosed by γ,\gamma, except possibly at a finite number of poles of ff in G. Let Z be the number of zeros of ff in G, and P the number of poles of ff in G, with each zero and pole counted as many times as its multiplicity. Show that 12πiγf(z)f(z)dz=ZP.\frac{1}{2 \pi i} \oint_{\gamma} \frac{f^{\prime}(z)}{f(z)} d z=Z-P. This formula is known as the argument principle.


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Answered 1 year ago
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Let f(z)f(z) have zeros z1,z2,,znz_1,z_2,\ldots,z_n of orders r1,r2,,rnr_1,r_2,\ldots,r_n respectively and have poles ξ1,ξ2,,ξm\xi_1,\xi_2,\ldots,\xi_m of orders s1,s2,,sms_1,s_2,\ldots, s_m respectively in GG so that

Z=r1+r2++rn  and  P=s1+s2++sm{\color{#4257b2}Z=r_1+r_2+\cdots+r_n}~~\text{and}~~{\color{#4257b2}P=s_1+s_2+\cdots+s_m}

Clearly, the only singularities of f(z)f(z)\frac{f'(z)}{f(z)} are z1,z2,,zn;ξ1,ξ2,,ξmz_1,z_2,\ldots,z_n;\xi_1,\xi_2,\ldots,\xi_m.

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