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Suppose \(f(z)\) is a meromorphic function (i.e., \(f(z)\) is analytic everywhere in the finite \(z\) plane except at isolated points where it has poles) with \(N\) simple zeroes (i.e., \(f\left(z_{0}\right)=0, f^{\prime}\left(z_{0}\right) \neq 0\) ) and \(M\) simple poles inside a circle \(C\). Show $$ \frac{1}{2 \pi i} \oint_{C} \frac{f^{\prime}(z)}{f(z)} d z=N-M $$

Short Answer

Expert verified
\( \frac{1}{2\text{π}i} \text{∮}_{C} \frac{f'(z)}{f(z)} \text{d}z = N - M \)

Step by step solution

01

Recall the Argument Principle

The Argument Principle states that if a meromorphic function has zeroes and poles inside a contour, the integral of its logarithmic derivative around the contour equals the difference between the number of zeros and poles.
02

Expressing the Principle Mathematically

The Argument Principle can be mathematically expressed as: \[ \frac{1}{2 \text{Ï€}i} \frac{\text{d}}{\text{d}z}\text{ln} f(z) = \text{no. of zeroes} - \text{no. of poles} \]
03

Differentiating the Logarithm

Using the chain rule, the above logarithm's derivative is computed as: \[ \frac{\text{d}}{\text{d}z} \text{ln} f(z) = \frac{f'(z)}{f(z)} \]
04

Integrating around the Contour

Integrate the derivative along the contour C: \[ \frac{1}{2 \text{π}i} \text{∮}_{C} \frac{f'(z)}{f(z)} \text{d}z = N - M \] where N represents the number of zeroes and M the number of poles.
05

Conclude the Relationship

The integral of \( \frac{f'(z)}{f(z)} \) around the closed contour C gives the difference between the number of zeroes (N) and the number of poles (M).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

meromorphic functions
A meromorphic function is a special type of function in complex analysis. It is analytic everywhere in the complex plane, except at isolated points where it has poles. A function is called analytic if it is differentiable at every point in its domain. For example, the function \(f(z) = \frac{1}{z}\) is meromorphic because it is analytic everywhere except at \(z = 0\), where it has a pole. Poles are specific types of singularities that indicate where the function goes to infinity.
In simpler terms, a meromorphic function behaves nicely almost everywhere, with only a few 'bad' points. These 'bad' points are just poles, and they are not complicated like other types of singularities.
These properties make meromorphic functions very useful in complex analysis, particularly when working with the Argument Principle, zeros, and poles.
logarithmic derivative
The logarithmic derivative of a function is used to understand how the function behaves around its zeros and poles. For a function \(f(z)\), its logarithmic derivative is defined as \(\frac{f'(z)}{f(z)}\).
This derivative leverages the natural logarithm of the function. If you recall from calculus, the derivative of \(\ln(f(z))\) is \(\frac{f'(z)}{f(z)}\). This is extremely useful because the logarithmic derivative simplifies the study of the function's growth and decay.
In the context of the Argument Principle, the logarithmic derivative helps us calculate the difference between the number of zeros and poles inside a given contour. This is pivotal in complex analysis and helps simplify otherwise complicated integrals.
zeros and poles
Zeros and poles are fundamental concepts in complex analysis which describe specific behaviors of a function.
A zero of a function \(f(z)\) is a point where \(f(z) = 0\). If \(f(z_0) = 0\) but \(f'(z_0) eq 0\), \(z_0\) is called a simple zero. A pole of a function, on the other hand, is a point where the function goes to infinity. If \(f(z)\) has a pole at \(z = p\), it means that \(f(z)\) behaves like \(\frac{1}{(z-p)^k}\) near \(p\) for some positive integer \(k\). If \(k = 1\), it is called a simple pole.
Understanding zeros and poles helps in evaluating integrals involving meromorphic functions and their logarithmic derivatives. They are key to applying the Argument Principle, as the count of zeros and poles within a contour gives valuable information about the function's behavior.
complex integration
Complex integration involves integrating complex-valued functions over contours in the complex plane. A contour is simply a smooth path along which we integrate.
The integral \(\oint_C f(z) \mathrm{d}z\) around a closed contour \(C\) is central to complex analysis. For meromorphic functions, these integrals are often evaluated using the Cauchy Integral Formula and the Residue Theorem.
The Argument Principle itself, involving complex integrals of logarithmic derivatives, states that the integral \(\oint_C \frac{f'(z)}{f(z)} \mathrm{d}z\) equals \(2\pi i \times (N - M)\). Here, \(N\) is the number of zeros, and \(M\) is the number of poles within the contour, highlighting the deep relationship between complex integration, zeros, and poles.

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Most popular questions from this chapter

Given the ODE $$ z y^{\prime \prime}+(2 r+1) y^{\prime}+z y=0 $$ look for a contour representation of the form \(y=\int_{C} e^{z \zeta} v(\zeta) d \zeta\). (a) Show that if \(C\) is a closed contour and \(v(\zeta)\) is single valued on this contour, then it follows that \(v(\zeta)=A\left(\zeta^{2}+1\right)^{r-1 / 2}\) (b) Show that if \(y=z^{-s} w\), then when \(s=r, w\) satisfies Bessel's equation \(z^{2} w^{\prime \prime}+z w^{\prime}+\left(z^{2}-r^{2}\right) w=0\), and a contour representation of the solution is given by $$ w=A z^{r} \oint_{C} e^{z \zeta}\left(\zeta^{2}+1\right)^{r-1 / 2} d \zeta $$ Note that if \(r=n+1 / 2\) for integer \(n\), then this representation yields the trivial solution. We take the branch cut to be inside the circle \(C\) when \((r-1 / 2) \neq\) integer.

The hypergeometric equation $$ z y^{\prime \prime}+(a-z) y^{\prime}-b y=0 $$ has a contour integral representation of the form \(y=\int_{C} e^{z \zeta} v(\zeta) d \zeta\). (a) Show that one solution is given by $$ y=\int_{0}^{1} e^{\tau 5} \zeta^{b-1}(1-\zeta)^{a-b-1} d \zeta $$ where \(\operatorname{Re} b>0\) and \(\operatorname{Re}(a-b)>0\) (b) Let \(b=1\), and \(a=2\); show that this solution is \(y=\left(e^{z}-1\right) /(z)\), and verify that it satisfies the equation. (c) Show that a second solution, \(y_{2}=v y_{1}\) (where the first solution is denoted as \(y_{1}\) ) obeys $$ z y_{1} v^{\prime \prime}+\left(2 z y_{1}^{\prime}+(a-z) y_{1}\right) v^{\prime}=0 $$ Integrate this equation to find \(v\), and thereby obtain a formal representation for \(y_{2}\). What can be said about the analytic behavior of \(y_{2}\), near \(z=0 ?\)

Show that $$ \int_{0}^{\infty} \frac{\cosh a x}{\cosh \pi x} d x=\frac{1}{2} \sec \left(\frac{a}{2}\right), \quad|a|<\pi $$ Use a rectangular contour with corners at \(\pm R\) and \(\pm R+i\).

Projection operators can be defined as follows. Consider a function \(F(z)\) $$ F(z)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta $$ where \(C\) is a contour, typically infinite (e.g. the real axis) or closed (e.g. a circle) and \(z\) lies off the contour. Then the "plus" and "minus" projections of \(F(z)\) at \(z=\zeta_{0}\) are defined by the following limit: $$ F^{\pm}\left(\zeta_{0}\right)=\lim _{z \rightarrow \zeta_{0}}\left[\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta\right] $$ where \(\zeta_{0}{ }^{\pm}\)are points just inside \((+)\)or outside (-) of a closed contour (i.e., \(\lim _{z \rightarrow 5_{0}+\text { denotes the limit from points } z \text { inside the contour } C \text { ) or to }}\) the left (t) or right (-) of an infinite contour. Note: the "+" region lies to the left of the contour; where we take the standard orientation for a contour, that is, the contour is taken with counterclockwise orientation. To simplify the analysis, we will assume that \(f(x)\) can be analytically extended in the neighborhood of the curve \(C\). (a) Show that $$ F^{\pm}\left(\zeta_{0}\right)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-\zeta_{0}} d \zeta \pm \frac{1}{2} f\left(\zeta_{0}\right) $$ where \(f_{C}\) is the principal value integral that omits the point \(\zeta=\zeta 0\). (b) Suppose that \(f(\zeta)=1 /\left(\zeta^{2}+1\right)\), and the contour \(C\) is the real axis (infinite), with orientation take from \(-\infty\) to \(\infty\); find \(F^{\pm}\left(\zeta_{0}\right)\). (c) Suppose that \(f(\zeta)=1 /\left(\zeta^{2}+a^{2}\right), a^{2}>1\), and the contour \(C\) is the unit circle centered at the origin with counterclockwise orientation; find \(F^{\pm}\left(\zeta_{0}\right)\)

(a) Use principal value integrals to show that $$ \int_{0}^{\infty} \frac{\cos k x-\cos m x}{x^{2}} d x=\frac{-\pi}{2}(|k|-|m|), \quad k, m \text { real. } $$ Hint: note that the function \(f(z)=\left(e^{i k z}-e^{i m z}\right) / z^{2}\) has a simple pole at the origin. (b) Let \(k=2, m=0\) to deduce that $$ \int_{0}^{\infty} \frac{\sin ^{2} x}{x^{2}} d x=\frac{\pi}{2} $$

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