Functions of a Complex Variable

Complex Functions

Example

\(w = f(z) = z^2 = f(x+yi) = (x^2-y^2) + 2xyi = u(x,y) + v(x,y)i\)

Definition

\(f(z) = u(x,y) + v(x,y)i\) where \(u(x,y)\) and \(v(x,y)\) are real functions.

Analytic Function

Cauchy Theorem

Classic Examples

Polynomials

\(p_n(z) = \sum_i a_iz^i\).

They are entire.

Power Series

\(f(z) = \sum_k^\infty a_k (z-z_0)^k\).

Holomorphic in radius of convergence, \(\mathcal{R}\). Typically convergent until your first nearest singularity.

Laurent Series

\(g(z) = \sum_{k=-\infty}^\infty a_k(z-z_0)^k = (z-z_0) + \sum_k^\infty a_k(z-z_0)^k + \frac{b_k}{(z-z_0)^k}\). If any \(b_k\neq 0\) then \(g(z)\) has a ``pole’’ at \(z_0\). The order of the pole is the largest \(k\) such that \(b_k\) is non-zero. I.e. \(1/z\) has a first order pole at \(z=0\) and \(1/(z+i)^2\) has a second order pole at \(-i\).

If \(f\) has a Laurent series, then it is holomorphic away from poles. - Meromorphic.

Pole Terminology

Order is largest \(k\) and first order is called a simple pole. Meromorphic with poles where \(p_m(z)\) has a root and \(p_n(z)\) does not have a root.

Rational Functions

\(f(z) = p_n(z)/p_m(z)\).

Exponential

\(\exp(z) = \sum_n \frac{1}{n!}z^n\) is entire.

Trigonometric

\(\sin(z),\sinh(z),\cos(z),\cosh(z),\tan(z),\tanh(z),\csc(z),\text{csch}(z),\sec(z),\text{sech}(z),\cot(z),\coth(z)\cdots\).

\(\sin,\cos,\cosh,\sinh\) are entire. (Might also have some others?)

Logarithmic

\(\ln z = u+iv\)

\(\exp(\log z) = z\), with \(z=r\exp(i\theta)\). So, \(z=\exp(u)\exp(iv) = \exp(u+iv)\), \(u=\ln r\). Then, \(v=\theta + 2\pi n\), \(n\in\mathbb{Z}\). So, \(\ln z\) is a multivalued function.

Multivalued Function and Branches

\(z=r\exp(i\theta)\). \(\ln z = \ln r + i(\arg z + 2\pi n)\).

Just like the square root gives two values, additional context lets you know which value is returned for that problem.

\(n=0\) gives the principal branch, Ln \(z\).

We can choose \(-\pi\lt\arg z\leq\pi\); \(n=0\), \(\pi\lt\arg z\leq 3\pi\); \(n=1\), \(-3\pi\lt\arg z\leq -\pi\); \(n=-1\).

In this way, we choose a branch cut from 0 to negative infinity along the negative real axis. Branch point, you cannot move a branch point from a branch cut.

Roots

\(z^{1/n}\)

\((z^{1/n})^n = z\).

\(z=r\exp(i\theta + 2n\pi)\). \(z^{1/n} = r^{1/n}\exp\frac{i\theta+2n\pi}{n}\).

So, \(z^{1/2} = \sqrt{r}\exp\frac{i\theta+2n\pi}{2} = \sqrt{r}\exp(i\theta/2)(-1)^n\). So, there are only 2 branches.

Derivatives

Derivatives of a complex function are the same as typical derivatives.

Contour Integrals

Motivation

Suppose we have two curves going from \(a\) to \(b\). If there are no singularities enclosed in the region bounded by the curves then the circulation is zero. If it contains a singularity, it is dependent on the singularities.

Cauchy-Gorsat Theorem

If \(\gamma\) is a piecewise regular closed contour and \(f(z)\) is holomorphic everywhere in and on \(\gamma\) then \(\int_\gamma dz f(z)=0\). Sometimes written \(\oint dz f(z)=0\).

Conventions

The integrals are typically oriented in a CCW fashion.

Cauchy Integral Representation

\(f(z) = \frac{1}{2\pi i}\oint_\gamma dz'f(z')/(z'-z)\) for \(z\) inside the contour \(\gamma\).

Proof. Because \(f(z)\) is holomorphic, we can deform the contour to anything around the point. So for instance, we can do a circle of radius \(\varepsilon\) around the point \(z\). For a small radius, \(z\to z'\). Then, \(f(z')=f(z)+f'(z')(z'-z)+\mathcal{O}(\varepsilon^2)\). So, \(\oint dz' \frac{f(z')}{z'-z} = \oint dz' \frac{f(z)+f'(z')(z'-z)}{z'-z} = \oint dz' \frac{f(z)}{z'-z} + f'(z') = 2\pi if(z) + 2\pi\varepsilon = 2\pi i f(z) + 0 = 2\pi if(z)\). Thus, we have \(f(z) = \frac{1}{2\pi i}2\pi if(z)=f(z)\).

Implications

Note, \(\frac{\partial^n}{\partial^n z}\frac{1}{z'-z} = (-1)^nn!\frac{(-1)^n}{(z'-z)^n} = \frac{n!}{(z'-z)^n}\). Thus, we have \(f^{(n)}(z) = \frac{n!}{2\pi i}\oint dz' \frac{f(z')}{(z'-z)^{n+1}}\)

Cauchy Residue Theorem Implications from Integral Representation

From this, we have, \(\oint dz' \frac{f(z')}{(z'-z)^{n+1}}=\frac{2\pi i}{n!}f^{(n)}(z)\)?

Aside

Notation: \(\partial\mathcal{D}\) denotes the boundary of \(\mathcal{D}\).

If \(f\) is holomorphic in a domain \(\mathcal{D}\), then \(|f(z)|\), \(\Re[f(z)]\), and \(\Im[f(z)]\) cannot have local extrema inside \(\mathcal{D}\) (i.e. their extrema are on \(\partial \mathcal{D}\)).

A bounded entire function is constant. Thus, all interesting functions either have singularities if they are bound at infinity or are unbound. Thus, a function is either constant or unbound at infinity.

Series Expansions

What about Regions in which \(f(x)\) is not Holomorphic everywhere?

I.e. where we have singularities.

Excise the singularity from the domain… Lauraunt expansion.

Laurent Series

Theorem

Suppose that \(f(z)\) is holomorphic in the annulus of \(\mathcal{C}_2\) and \(\mathcal{C}_1\) which are circles with the positive orientation (CCW) centered around \(z_0\). Then there exists a series: \(f(z) = \sum_{n=-\infty}^\infty d_n(z-z_0)^n,d_n=\frac{1}{2\pi i}\oint_\gamma dz' \frac{f(z')}{(z'-z_0)^{n+1}}\) that converges uniformily to \(f(z)\) inside the annulus. \(\gamma\) is any contour inside the annulus.

This is also written as \(f(z) = \sum_{n=1}^{\infty} \frac{b_n}{(z-z_0)^n} + \sum_{m=0}^\infty a_m(z-z_0)^m, a_n=\frac{1}{2\pi i}\oint dz' \frac{f(z')}{(z'-z_0)^{n+1}}, b_n = \frac{1}{2\pi i}\oint dz' (z-z_0)^{n-1}f(z')\).

The \(b_n\) series is called the principal part.

Proof. Similar to Taylor series. Let \(\mathcal{C}_2\) be the outer circle. Then we can go along \(\mathcal{C}_2\) and as we get close to the beginning, go down to \(\mathcal{C}_1\), then traverse \(\mathcal{C}_1\) backwards, as we get close to the beginning of \(\mathcal{C}_1\) we jump back up to \(\mathcal{C}_2\)’s beginning.

Finding Coefficients

Calculating Laurent series coefficients is typically done by: guess and check, partial fractions, leverage the binomial expansion, taylor series.

Example

\(f(z)=z\exp(-1/z)\). Expect: \(z=0\). Expand with \(z_0=0\). Then, \(f(z) = z\sum_n \frac{1}{n!z^n} = \frac{(-1)^n}{n!z^{n-1}}\). The principal part goes from \(n=2\) onwards. \(a_1 = 1\), \(a_0=-1\), \(b_n=\frac{(-1)^(n+1)}{(n+1)!}\), \(\cdots\).

Classifying Singularities

Cauchy Residues

Examples

Example 1

\(f(z)=\frac{1}{z}\).

So, Res \(f(0) = \lim_{z\to0}(z)f(z) = \frac{z}{z} = 1\).

Example 2

\(f(z)=\frac{1}{z^2}\). So, Res \(f(0) = \lim_{z\to 0}\frac{d}{dz}z^2f(z) = \lim_{z\to 0}\frac{d}{dz}1 = 0\).

Example 3

\(f(z) = \frac{z^2+5z+3}{(z-1)(z+2)^2}\). Singularities at \(z=1\) and \(z=-2\).

Res \(f(1) = \lim_{z\to 1} \frac{z^2 + 5z+3}{(z+2)^2} = \frac{1 + 5 + 3}{9} = 1\).

Res \(f(-2) = \lim_{z\to -2} \frac{d}{dz}\frac{z^2+5z+3}{z-1} = \lim_{z\to -2} \frac{(2z+5)(z-1) - (z^2+5z+3)(1)}{(z-1)^2} = \frac{(-4+5)(-3) - (4-10+3)}{9} = \frac{-3+3}{9} = 0\).

Example 4

\(h(z) = \frac{z^2}{z^2 + 3} = \frac{z^2}{(z+3i)(z-3i)}\) Singularities at \(\pm 3i\).

Then, Res \(f(\sqrt{3}i) = \frac{(\sqrt{3}i)^2}{(\sqrt{3}i+\sqrt{3}i)} = \frac{\sqrt{3}}{2}i\).

Res \(f(-\sqrt{3}i) = \frac{\sqrt{3}}{2i}\).

Example 5

\(I=\int_0^\infty dx \frac{x^2}{(x^2+1)(x^2+3)} = \int_0^\infty dx \frac{x^2}{(x+i)(x-i)(x+\sqrt{3}i)(x-\sqrt{3}i)}\). This function is even. So, \(2I=\int_{-\infty}^\infty dx \frac{x^2}{(x^2+1)(x^2+3)}\). Since this is a rational function, \(\int_{-\infty}^\infty dx f(z) = 2\pi i\text{Res }f(-i) + 2\pi i\text{Res }f(\sqrt{3}i)\) These are simple poles, with residues: Res \(f(i) = \frac{(i)^2}{d_x(x^4+4x^2+3)} = \frac{-1}{-4(i)+8i}=\frac{i}{4}\).

Res \(f(\sqrt{3}i)=\frac{3(i)^2}{-4i\sqrt{3}^3+8i\sqrt{3}}=\frac{-3(-i)}{\sqrt{3}(8-4)i(-i)}=\frac{3i}{4\sqrt{3}}\).

\(I=\pi i(\frac{1}{4}-\frac{\sqrt{3}}{4})i=\frac{\pi}{4}(\sqrt{3}-1)\).

Example 6

An integral that arises in optics.

\(I=\int_0^\infty dx\frac{\sin x}{x}=\frac{1}{2}\int_{-\infty}^\infty dx\frac{\sin x}{x}=\frac{1}{4i}\int_{-\infty}^\infty\frac{\exp(ix)-\exp(-ix)}{x} = \frac{1}{2}\Im\int_{-\infty}^\infty\frac{\exp ix}{x}\). This has a simple pole at 0.

The exponential approaches zero in the upper half of the plane but explodes in the bottom half of the plane. Thus, we see that the top semicircle contributes nothing and closes the contour.

Our integral traverses through the singularity - our theorems don’t apply. So the answer is ambiguous. We may choose the path \(\Gamma_\varepsilon^-\) or \(\Gamma_\varepsilon^+\).

\(I = \frac{1}{2}\Im\lim_{R\to\infty}\{\int_{-R}^{R}+\int_{\Gamma_R^+}dzf(z)+\cdot\}\). In this case, both \(\Gamma_\varepsilon\) paths yield the same result. \(I = \frac{1}{2}\Im\lim_{R\to\infty,\varepsilon\to0}\{\int_{-R}^{-\varepsilon}f(x)dx+\int_{\varepsilon}^Rf(x)dx+\int_{\Gamma_R^+}dzf(z)+\int_{\Gamma_\varepsilon}f(z)dz\}\).

\(\int_{\gamma_\varepsilon^+}\frac{\exp(iz)}{z}=-\pi i\)

For the top path, we see that we are holomorphic in the entire region, thus we get 0. For the bottom path, we have a simple pole and in the limit of z to zero, thus we have zero. Thus, either way we find the overall contour integral to be zero. Therefore, \(\int_{-\infty}^\infty \frac{\exp iz}{z}dz=\pi i\). Thus, our original integral is \(\frac{\pi}{2}\).

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:24

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