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Question Number 213862 by issac last updated on 19/Nov/24
Help me.....!!!  :(    complex anaylsis problem..  f(z) is entire in path C   entire: Differantiable complex function  mean f(z) satisfy f(z)=u(x,y)+i∙v(x,y)    (∂u/∂x)=−(∂v/∂y) or  (∂u/∂y)=−(∂v/∂x) (couchy-riemann)  show that ∫_( C)  ((f(z))/(f′(z))) dz=2πiΣ_(h=1) ^M  P_h −Q_h   P_h  is number of Zeros in path C   Q_h  is number of poles in path C  and if pole is not exist  show that (1/(2πi)) ∫_( C)  ((f(z))/(f′(z))) dz   this equation  is equivalent to the   formula for finding the number of zeros  f(z)=0
$$\mathrm{Help}\:\mathrm{me}…..!!!\:\::\left(\:\:\right. \\ $$$$\mathrm{complex}\:\mathrm{anaylsis}\:\mathrm{problem}.. \\ $$$${f}\left({z}\right)\:\mathrm{is}\:\mathrm{entire}\:\mathrm{in}\:\mathrm{path}\:{C}\: \\ $$$$\mathrm{entire}:\:\mathrm{Differantiable}\:\mathrm{complex}\:\mathrm{function} \\ $$$$\mathrm{mean}\:{f}\left({z}\right)\:\mathrm{satisfy}\:{f}\left({z}\right)={u}\left({x},{y}\right)+\boldsymbol{{i}}\centerdot{v}\left({x},{y}\right)\:\: \\ $$$$\frac{\partial{u}}{\partial{x}}=−\frac{\partial{v}}{\partial{y}}\:\mathrm{or}\:\:\frac{\partial{u}}{\partial{y}}=−\frac{\partial{v}}{\partial{x}}\:\left(\mathrm{couchy}-\mathrm{riemann}\right) \\ $$$$\mathrm{show}\:\mathrm{that}\:\int_{\:{C}} \:\frac{{f}\left({z}\right)}{{f}'\left({z}\right)}\:\mathrm{d}{z}=\mathrm{2}\pi\boldsymbol{{i}}\underset{{h}=\mathrm{1}} {\overset{{M}} {\sum}}\:{P}_{{h}} −{Q}_{{h}} \\ $$$${P}_{{h}} \:\mathrm{is}\:\mathrm{number}\:\mathrm{of}\:\mathrm{Zeros}\:\mathrm{in}\:\mathrm{path}\:{C}\: \\ $$$${Q}_{{h}} \:\mathrm{is}\:\mathrm{number}\:\mathrm{of}\:\mathrm{poles}\:\mathrm{in}\:\mathrm{path}\:{C} \\ $$$$\mathrm{and}\:\mathrm{if}\:\mathrm{pole}\:\mathrm{is}\:\mathrm{not}\:\mathrm{exist} \\ $$$$\mathrm{show}\:\mathrm{that}\:\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{\:{C}} \:\frac{{f}\left({z}\right)}{{f}'\left({z}\right)}\:\mathrm{d}{z}\: \\ $$$$\mathrm{this}\:\mathrm{equation}\:\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{formula}\:\mathrm{for}\:\mathrm{finding}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{zeros} \\ $$$${f}\left({z}\right)=\mathrm{0} \\ $$

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