Cauchy's Theorem. The Argument Principle

Put

Compute the sum

which converges to the integral $\oint \dfrac {dz}{z}$ along $\left|z\right|=\left|a\right|$ as $n\to \infty$.

Let $k$ denote an integer different from $-1$ and $L$ a closed curve without double points and of finite length; if $k\leq -2$, $L$ does not pass through $z=0$; furthermore the points $z_{1},z_{2},\ldots ,z_{n}$ are consecutive points on $L$. Show that the integral $\oint_{L} z^{k}\, dz$ vanishes; try to approximate it by a sum of the form

Evaluate

along the ellipse $z=x+iy$, $x^{2}-xy+y^{2}+x+y=0$.

Show that

Prove the formula

Evaluate the curvilinear integral

along an ellipse with the foci $(0,-1)$ and $(0,1)$.

We have for $0<\mathrm{Re}\left(s\right)<1$

The equation

holds for $n>1$.

We find, assuming $\mu >0$, $0<\alpha <\frac{\pi }{2}$, $n=0,1,2,\ldots$, that

Note the special case $\alpha =\mu \pi$.

Assume $\mu >0$, $x>0$, $\mu$ fixed, $x$ variable. Show that

Let $a>0$. The integral

along the straight line $s=a+it$, $-\infty <t<\infty$, parallel to the imaginary axis converges absolutely for all real values of $\alpha$. The integral turns out to be

We call $\mu (t)$ the largest term of the series

Assume $\lambda >0$ and let $z$ be the only positive root of the equation

Then

The Legendre polynomials can be defined as the coefficients of the expansion into a power series [Part VI, problem 91]

Deduce from this Laplace's formula (Part VI, problem 86)

and the Dirichlet-Mehler formula

(The square roots are positive.)

Let $\mathfrak {H}$ denote the half-strip

and $L$ the boundary of $\mathfrak {H}$ consisting of three straight pieces. We orient $L$ so that $\mathfrak {H}$ is on the right hand side of $L$. The integral

defines a function $E(z)$ for points $z$ on the left hand side of $L$. Show that $E(z)$ is an entire function which assumes real values for real $z$.

(continued). We find

(continued). The function

is bounded outside $\mathfrak {H}$. Inside $\mathfrak {H}$ the function

is bounded.

We define

Show that

where the integrals are computed along the positively oriented circles. [Part II, problem 217.]

Assume that $f(z)$ is regular in the disk $\left|z\right|\leq r$ and different from $0$ on the circle $\left|z\right|=r$. The largest value of $\re z\dfrac {f’(z)}{f(z)}$ on $\left|z\right|=r$ is at least equal to the number of zeros of $f(z)$ in $\left|z\right|<r$.

Let $z_{1},z_{2},\ldots ,z_{n}$ be arbitrary but distinct complex numbers and $L$ be a closed continuous curve without double points, enclosing all the $z_{\nu }$'s. The function $f(z)$ is supposed to be regular inside of, and on, $L$. Then

where $\omega (z)=(z-z_{1})(z-z_{2})\cdots (z-z_{n})$, is the uniquely determined polynomial of degree $n-1$ that coincides with $f(z)$ at the points $z_{1},z_{2},\ldots ,z_{n}$.

The function $f(z)$ is analytic on the segment $a\leq z\leq b$ of the real axis and it assumes there real values. The closed curve $L$ is continuous without double points and encloses the segment $a\leq z\leq b$; $f(z)$ is regular inside $L$. Let $z_{1},z_{2},\ldots ,z_{n}$ denote arbitrary points of the real segment $[a,b]$. Then there exists a point $z_{0}$ on $[a,b]$ such that

Assume that the entire function $F(z)$ satisfies the inequality

in the entire $z$-plane, $z=x+iy$; $C$ and $\varrho$ are positive constants. Then

Example: $F(z)=\cos \varrho z$.

The entire function $G(z)$ is supposed to satisfy the inequality

on the entire $z$-plane, $z=x+iy$; $C,\varrho$ positive constants; in addition assume that $G(z)$ is an odd function, $G(-z)=-G(z)$. Then

Example: $G(z)=\sin \varrho z$.

We suppose that the function $f(z)$ is regular in the disk $\left|z\right|\leq 1$. Then

We integrate with respect to $x$ along a straight line from $0$ to $1$, with respect to $z$ along the positively oriented unit circle. We start at the point $z=1$ with the branch of $\log z$ that is real, and the branch of $z^{k}$ that is positive, for positive $z$.

The function $f(z)$ is regular in the unit disk $\left|z\right|\leq 1$ and satisfies the condition

Let $k>-1$. Then

Let $\alpha >-2$. The quadratic form of the infinitely many real variables $x_{1},x_{2},x_{3},\ldots$

is bounded, i.e.there exists a constant $M$ independent of $n$ such that

whenever the variables $x_{1},x_{2},\ldots ,x_{n}$ satisfy the condition $x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}=1$; $n=1,2,3,\ldots$ We may choose $M=\pi$ if $\alpha$ is an integer and $M=\dfrac {\pi }{\left|\sin \alpha \pi \right|}$ if $\alpha$ is not an integer.

We assume that the functions $f_{1}(z),f_{2}(z),\ldots ,f_{n}(z),\ldots$ are regular in the open region $\mathfrak {R}$ and that they converge uniformly to the function $f(z)$ in any closed domain inside $\mathfrak {R}$. Then the limit function $f(z)$ is regular in $\mathfrak {R}$.

The complex function $f(z)=u(x,y)+iv(x,y)$ of the real variables $x$ and $y$ is defined and continuous in a region $\mathfrak {R}$ of the $z=x+iy$ plane. Moreover we assume that the integral

along any circle inside $\mathfrak {R}$ vanishes. Then $f(z)$ is an analytic function of the complex variable $z$, regular in the entire region $\mathfrak {R}$.

The function $f(z)$ is assumed to be analytic in the open disk $\left|z\right|<1$, bounded on the closed disk $\left|z\right|\leq 1$ and continuous, possibly with the exception of a finite number of points. Then

(More general than problem 118.)

The function $f(z)$ is regular in the disk $\left|z\right|\leq R$; let $0<r<R$. Then we have Poisson's formula

Suppose that the function $f(z)$ is regular and bounded in the half-plane $\mathrm{Re}\left(z\right)\geq 0$. Then, for $x>0$,

We assume that the function $f(z)$ is meromorphic in the disk $\left|z\right|\leq 1$, regular and non-zero on the boundary and at the origin. The zeros of $f(z)$ in $\left|z\right|\leq 1$ are $a_{1},a_{2},\ldots ,a_{m}$ and the poles $b_{1},b_{2},\ldots ,b_{n}$ (multiple zeros and poles are listed with correct multiplicity). Then we have Jensen's formula

The function $f(z)$ is meromorphic in the disk $\left|z\right|\leq R$, regular and different from $0$ on the circle $\left|z\right|=R$ and it has inside the circle the zeros $a_{1},a_{2},\ldots ,a_{m}$ and the poles $b_{1},b_{2},\ldots ,b_{n}$, counted with correct multiplicities. If the point $z=re^{i\vartheta }$, $r<R$, is neither a zero nor a pole we have

The function $f(z)$ is meromorphic in the half-plane $\mathrm{Re}\left(z\right)\geq 0$, regular and non-zero on its boundary and it has inside the half-plane the zeros $a_{1},a_{2},\ldots ,a_{m}$ and the poles $b_{1},b_{2},\ldots ,b_{n}$, counted with correct multiplicity. If $f(z)$ is regular at infinity (but also under weaker conditions on its behaviour at infinity) and if $f(z)$ is regular and non-zero at $z=x+iy$, $x>0$, we have

The function $f(z)$ is regular in the domain

different from zero on the boundary of $\mathfrak {D}$ and it has in $\mathfrak {D}$ the zeros $a_{1},a_{2},\ldots ,a_{m}$, $a_{\mu }=r_{\mu }e^{i\vartheta_{\mu }}$, $\mu =1,2,\ldots ,m$. Using the definition $\log \left|f(\varrho e^{i\vartheta })\right|=U(\varrho ,\vartheta )$ we have the formula

where $\chi (R)$ is bounded as $R\to \infty$; $r,f(z)$ are fixed.

Prove problem 25 by examining the change of $\arctan \dfrac {V(x)}{U(x)}$ as $x$ increases from $-\infty$ to $+\infty$ along the real axis.

We consider a closed, continuous, oriented curve in the $z$-plane that avoids the origin. If, starting from an arbitrary point, $z$ describes the entire curve in the given direction (returning to its starting point) the argument of $z$ changes continuously and its total variation is a multiple, $2\pi n$, of $2\pi$. The integer $n$ is called the winding number of the curve.

Every ray from the origin intersects the curve in question at least $\left|n\right|$ times.

Here $L$ denotes a closed continuous curve without double points and $\mathfrak {D}$ the closed interior of $L$. The winding number of a closed, continuous, oriented curve avoiding the origin is the integer $n$ such that the argument of $w$ changes by $2\pi n$ as $w$ describes the entire curve once in the given direction.

The functions $\varphi (z)$ and $\psi (z)$ are regular in $\mathfrak {D}$, except possibly at finitely many poles, finite and different from $0$ on the boundary $L$ of $\mathfrak {D}$. We define $f(z)=\varphi (z)\psi (z)$. The winding number of the image under $w=f(z)$ of $L$ is equal to the sum of the winding numbers of the images of $L$ under $w=\varphi (z)$ and $w=\psi (z)$ respectively.

Let $L$ denote a closed continuous curve without double points and $\mathfrak {D}$ the closed interior of $L$. The argument principle states: if $f(z)$ is regular in $\mathfrak {D}$, except possibly at finitely many poles, finite and non-zero on $L$, then as $z$ moves along $L$ in the positive sense the point $w=f(z)$ describes a closed continuous curve the winding number of which is equal to the number of zeros inside $L$ minus the number of poles inside $L$.

Prove the argument principle for a polynomial.

Let $L$ denote a closed continuous curve without double points and $\mathfrak {D}$ the closed interior of $L$. The argument principle states: if $f(z)$ is regular in $\mathfrak {D}$, except possibly at finitely many poles, finite and non-zero on $L$, then as $z$ moves along $L$ in the positive sense the point $w=f(z)$ describes a closed continuous curve the winding number of which is equal to the number of zeros inside $L$ minus the number of poles inside $L$.

The argument principle implies: If $\varphi (z)$ is regular and non-zero in the domain $\mathfrak {D}$ the winding number of the image of $L$ (the boundary curve of $\mathfrak {D}$) generated by $w=\varphi (z)$ vanishes; i.e.the argument of $\varphi (z)$ is a single-valued function on $L$. Deduce the general statement of the principle from this particular case.

The real trigonometric polynomial

has at least $2m$ and at most $2n$ zeros in the interval $0\leq \vartheta <2\pi$.

Let $0<a_{1}<a_{2}<\cdots <a_{n}$. Then the trigonometric polynomial

has $2n$ distinct zeros in the interval $0\leq \vartheta <2\pi$. [Problem 22.] [Consequently it has real zeros only, Part VI, problem 14.]

Here $L$ denotes a closed continuous curve without double points.

The function $f(z)$ is meromorphic in the interior of the curve $L$ and regular on the curve $L$. If $\left|a\right|$ is larger than the maximum of $\left|f(z)\right|$ on $L$ then $f(z)$ assumes the value $a$ inside $L$ just as often as it has poles there.

The function

assumes any value $w$ with positive real part once and only once in the half-strip $\mathrm{Re}\left(z\right)>0$, $-\frac{1}{2}<\mathrm{Im}\left(z\right)<\frac{1}{2}$.

Suppose that $f(z)$ is regular in the closed disk $\left|z\right|\leq r$ and univalent on the circle $\left|z\right|=r$, that is it assumes there no value more than once. Then the image of $\left|z\right|=r$ under $w=f(z)$ has the same orientation as $\left|z\right|=r$ and the function $f(z)$ is univalent (schlicht) also in the disk $\left|z\right|<r$.

The zeros of the function $\displaystyle \int_{0}^{z} e^{-\frac{x^{2}}{2}}\, dx$ lie with the exception of $z=0$ inside the region $\mathrm{Re}\left(z^{2}\right)<0$. [Cornu spiral, cf.e.g.A. Sommerfeld: Optics. New York: Academic Press 1954, pp.243--244].

The winding number of a closed, continuous, oriented curve avoiding the origin is the integer $n$ such that the argument of $w$ changes by $2\pi n$ as $w$ describes the entire curve once in the given direction.

The function $f(z)$ is single-valued, regular and does not assume a certain value $a$ in the annulus $r<\left|z\right|<R$. All the closed continuous curves without double points that enclose the circle $\left|z\right|=r$ and that lie inside the annulus are mapped by the function $w=f(z)-a$ onto curves in the $w$-plane with the same winding number.

Here $L$ denotes a closed continuous curve without double points and $\mathfrak {D}$ the closed interior of $L$.

The function $f(z)$ is regular in the domain $\mathfrak {D}$ and its absolute value on the boundary curve $L$ of $\mathfrak {D}$ is constant. As $z$ moves on the curve $L$ the argument of $f(z)$ changes monotonically. (Whence a new proof of problem 142.)

Here $L$ denotes a closed continuous curve without double points and $\mathfrak {D}$ the closed interior of $L$.

The function $f(z)$ has one zero more than $f'(z)$ inside $L$ under the hypothesis of problem 191. (More informative than problem 142.) This means geometrically: Inside a closed level line without double points the modular graph has more pits than saddle points, namely precisely one more.

Here $L$ denotes a closed continuous curve without double points and $\mathfrak {D}$ the closed interior of $L$.

If $f(z)$ is regular in the domain $\mathfrak {D}$ and $f'(z)$ does not vanish in $\mathfrak {D}$ the mapping $w=f(z)$ of $\mathfrak {D}$ is not necessarily schlicht [problem 72]. If however $\left|f(z)\right|$ is constant on the boundary of $\mathfrak {D}$ the mapping in question has to be schlicht.

Here $L$ denotes a closed continuous curve without double points and $\mathfrak {D}$ the closed interior of $L$.

We suppose that $f(z)$ and $\varphi (z)$ are two functions that are regular in the interior of $\mathfrak {D}$, continuous in the closed domain $\mathfrak {D}$ and that furthermore $\left|f(z)\right|>\left|\varphi (z)\right|$ on the boundary $L$ of $\mathfrak {D}$. Then the function $f(z)+\varphi (z)$ has exactly the same number of zeros inside $\mathfrak {D}$ as $f(z)$.

Let $\lambda$ be real, $\lambda >1$. The equation

has exactly one root in the disk $\left|z\right|=1$. This root is real and positive.

Let $\lambda$ be real, $\lambda >1$. The equation

has only one root in the half-plane $\mathrm{Re}\left(z\right)\geq 0$ which, consequently, is real.

A function (not necessarily schlicht) that maps the closed unit disk onto a domain contained in the open unit disk has exactly one fixed point. I.e.if $f(z)$ is regular in the disk $\left|z\right|\leq 1$ and if $\left|f(z)\right|<1$ in $\left|z\right|\leq 1$ then the equation $f(z)-z=0$ has exactly one root in $\left|z\right|\leq 1$.

The entire function $\dfrac {1}{\Gamma (z)}$ assumes each value infinitely often in the half-strip

Let $f(t)$ be a real-valued twice continuously differentiable function on the interval $0\leq t\leq 1$. If $\left|f(1)\right|>\left|f(0)\right|$ the entire function

has infinitely many real zeros and only a finite number of complex zeros; if $0<\left|f(1)\right|<\left|f(0)\right|$ it has only a finite number of real, and infinitely many complex, zeros. [The zeros of $F(z)$ behave with respect to reality like the zeros of $f(0)-f(1)\cos z$.]

Let $a$ be a constant, $\left|a\right|>2.5$. The power series

defines an entire function which does not vanish on the boundary of the annulus

and has exactly one zero inside the annulus, $n=1,2,\ldots$. [Examine the maximum term on the circle $\left|z\right|=\left|a\right|^{2n}$, problem 117.]

(continuation of problem 170). Let $\mathfrak {S}$ denote the set of all zeros of all the functions $f_{n}(z)$, $n=1,2,3,\ldots$ in $\mathfrak {R}$. If the limit function does not vanish identically its zeros in $\mathfrak {R}$ are identical with the limit points of $\mathfrak {S}$ in $\mathfrak {R}$. (The term ``limit point'' is used here to mean a point an arbitrary fixed neighborhood of which contains at least one zero of $f_{n}(z)$ for all sufficiently large $n$.)

The functions $f_{1}(z),f_{2}(z),\ldots ,f_{n}(z),\ldots$ are schlicht in the unit disk $\left|z\right|<1$ and converge in any smaller disk $\left|z\right|\leq r<1$ uniformly to a not everywhere constant limit function $f(z)$. Then the function $f(z)$ is schlicht in the unit disk $\left|z\right|<1$.