Laurent Series with WolframAlpha

WolframAlpha can compute (simple) Laurent series:

Series[Sin[z^(-1)], {z, 0, 5}]

1/z-1/(6 z^3)+1/(120 z^5)+O((1/z)^6)
(Laurent series)
(converges everywhere away from origin)

Unfortunately, more “complex” (pun intended) Laurent series are not possible for WolframAlpha.

A holomorphic and injective function has nonzero derivative

This post proves that if f:U\to V is a function that is holomorphic (analytic) and injective, then f'(z)\neq 0 for all z in U. The condition of having nonzero derivative is equivalent to the condition of conformal (preserves angles). Hence, this result can be stated as “A holomoprhic and injective function is conformal.”

(Proof modified from Stein-Shakarchi Complex Analysis)

We prove by contradiction. Suppose to the contrary f'(z_0)=0 for some z_0\in D. Using Taylor series, \displaystyle f(z)=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\dots

Since f'(z_0)=0, \displaystyle f(z)-f(z_0)=a(z-z_0)^k+G(z) for all z near z_0, with a\neq 0, k\geq 2 and G(z)=(z-z_0)^{k+1}H(z) where H is analytic.

For sufficiently small w\neq 0, we write \displaystyle f(z)-f(z_0)-w=F(z)+G(z), where F(z)=a(z-z_0)^k-w.

Since |G(z)|<|F(z)| on a small circle centered at z_0, and F has at least two zeroes inside that circle, Rouche’s theorem implies that f(z)-f(z_0)-w has at least two zeroes there.

Since the zeroes of a non-constant holomorphic function are isolated, f'(z)\neq 0 for all z\neq z_0 but sufficiently close to z_0.

Let z_1, z_2 be the two roots of f(z)-f(z_0)-w. Note that since w\neq 0, z_1\neq z_0, z_2\neq z_0. If z_1=z_2, then f(z)-f(z_0)-w=(z-z_1)^2h(z) for some analytic function h. This means f'(z_0)=0 which is a contradiction.

Thus z_1\neq z_2, which implies that f is not injective.

Rouche’s Theorem

Rouche’s Theorem

If the complex-valued functions f and g are holomorphic inside and on some closed contour K, with |g(z)|<|f(z)| on K, then f and f+g have the same number of zeroes inside K, where each zero is counted as many times as its multiplicity.


Consider the polynomial z^5+3z^3+7 in the disk |z|<2. Let g(z)=3z^3+7, f(z)=z^5, then

\begin{aligned}  |3z^3+7|&<3(8)+7\\  &=31\\  &<32\\  &=|z^5|  \end{aligned}
for every |z|=2.
Then f+g has the same number of zeroes as f(z)=z^5 in the disk |z|<2, which is exactly 5 zeroes.

Cauchy-Riemann Equations

Cauchy-Riemann Equations

Let f(x+iy)=u(x,y)+iv(x,y). The Cauchy-Riemann equations are:

\begin{aligned}  u_x&=v_y\\  u_y&=-v_x.  \end{aligned}

Alternative Form (Wirtinger Derivative)

The Cauchy-Riemann equations can be written as a single equation \displaystyle \frac{\partial f}{\partial\bar z}=0 where \displaystyle \frac{\partial}{\partial\bar z}=\frac 12(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}) is the Wirtinger derivative with respect to the conjugate variable.

Goursat’s Theorem

Suppose f=u+iv is a complex-valued function which is differentiable as a function f:\mathbb{R}^2\to\mathbb{R}^2. Then f is analytic in an open complex domain \Omega iff it satisfies the Cauchy-Riemann equations in the domain.

dz and dz bar: How to derive the Wirtinger derivatives

Something interesting in Complex Analysis is the Wirtinger derivatives:

\displaystyle\boxed{\frac{\partial}{\partial z}:=\frac 12(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})}

\displaystyle\boxed{\frac{\partial}{\partial \bar z}:=\frac 12(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})}

They are often simply defined as such, but one would be curious how to derive them, at least heuristically.

How to derive Wirtinger derivatives

It turns out we can derive them as such. Any complex function f(z) can be viewed as a function f(x,y) by considering z=x+iy. Since x=\frac 12 (z+\bar z), y=-\frac 12 i(z-\bar z), we can also view f(x,y) as f(z,\bar z).

Then by the Chain Rule (for multivariable calculus), we have \displaystyle\frac{\partial}{\partial x}=\frac{\partial z}{\partial x}\frac{\partial}{\partial z}+\frac{\partial\bar z}{\partial x}\frac{\partial}{\partial\bar z}=\frac{\partial}{\partial z}+\frac{\partial}{\partial\bar z}.

Similarly, we get \displaystyle\frac{\partial}{\partial y}=i(\frac{\partial}{\partial z}-\frac{\partial}{\partial\bar z}).

Then, solving the simultaneous equations we get the Wirtinger derivatives.

\displaystyle i\frac{\partial}{\partial x}+\frac{\partial}{\partial y}=2i\frac{\partial}{\partial z}. Thus, \displaystyle\frac{\partial}{\partial z}=\frac 12(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}).

Similarly, we can get that \displaystyle\boxed{\frac{\partial}{\partial \bar z}:=\frac 12(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})}.

Using Wirtinger derivatives, we can express the Cauchy-Riemann equations in a succinct manner: A function satisfies the Cauchy-Riemann equations iff \displaystyle\frac{\partial f}{\partial\bar z}=0.

Schwarz Lemma & Maximum Modulus Principle

Schwarz Lemma

Let D=\{z:|z|<1\} be the open unit disk in the complex plane \mathbb{C} centered at the origin and let f:D\to D be a holomorphic map such that f(0)=0.

Then, |f(z)|\leq |z| for all z\in D and |f'(0)\leq 1.

Moreover, if |f(z)|=|z| for some non-zero z or |f'(0)|=1, then f(z)=az for some a\in\mathbb{C} with |a|=1.

Maximum modulus principle

Let f be a function holomorphic on some connected open subset D of the complex plane \mathbb{C} and taking complex values. If z_0 is a point in D such that |f(z_0)|\geq|f(z)| for all z in a neighborhood of z_0, then the function f is constant on D.

Informally, the modulus |f| cannot exhibit a true local maximum that is properly within the domain of f.

Evaluation of Improper Integral via Complex Analysis

We are following the notation in Complex Variables and Applications (Brown and Churchill).

The method of using complex analysis to evaluate integrals is to consider a very large semicircular region’s boundary, which consists of the segment of the real axis from z=-R to z=R and the top half of the circle |z|=R positively oriented is denoted by C_R.

\int_{-R}^R f(x)\,dx+\int_{C_R}f(z)\,dz=2\pi i\sum_{k=1}^n\text{Res}_{z=z_k}f(z). If \lim_{R\to\infty}\int_{C_R}f(z)\,dz=0, then P.V.\int_{-\infty}^\infty f(x)\,dx=2\pi i\sum_{k=1}^n\text{Res}_{z=z_k}f(z). Furthermore if f is even, then \int_0^\infty f(x)\,dx=\pi i\sum_{k=1}^n\text{Res}_{z=z_k}f(z).

Useful Theorem

Let two functions p and q be analytic at a point z_0. If p(z_0)\neq 0, q(z_0)=0, and q'(z_0)\neq 0, then z_0 is a simple pole of the quotient p(z)/q(z) and \text{Res}_{z=z_0}\frac{p(z)}{q(z)}=\frac{p(z_0)}{q'(z_0)}.

Evaluating Integrals using Complex Analysis

Our next few posts on complex analysis will focus on evaluating real integrals like \displaystyle\int_0^\infty \frac{1}{x^2+1}\,dx using residue theory from Complex Analysis. This is something amazing about Complex Analysis, it can be used to solve integrals in real numbers, something which is not immediately obvious.

To calculate those real integrals, the first step is to study the theory of residues and poles. This can be found in Chapter 6 of Churchill’s book Complex Variables and Applications (Brown and Churchill).

The extremely powerful theorem that one first needs to know is called Cauchy’s Residue Theorem:

Let C be a simple closed positively oriented contour. If a function f is analytic inside and on C except for a finite number of singular points z_k (k=1,2,\dots,n) inside C, then \displaystyle \int_Cf(z)\,dz=2\pi i\sum_{k=1}^n\text{Res}_{z=z_k}\,f(z).

A Summary of the 3 types of Isolated Singular Points:

  1. Pole of order m. The coefficients b_n of the Laurent series contain a finite (nonzero) number of nonzero terms, i.e. b_n eventually becomes zero after a certain number. i.e. \displaystyle f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^n}+\dots+\frac{b_m}{(z-z_0)^m}.
  2. Removable singular point. Every b_n is zero.
  3. Essential singular point. An infinite number of the coefficients b_n in the principal part are nonzero.

Shortcut for calculating Residues at Poles

Theorem: An isolated singular point z_0 of a function f is a pole of order m if and only if f(z) can be written in the form f(z)=\frac{\phi(z)}{(z-z_0)^m} where \phi(z) is analytic and nonzero at z_0. Moreover \text{Res}_{z=z_0}f(z)=\phi(z_0) if m=1 and \text{Res}_{z=z_0}f(z)=\frac{\phi^(m-1)(z_0)}{(m-1)!}. if m\geq 2.

This is a very short crash course on the theorems needed. The next blog post on complex analysis will go into calculating some actual integrals.

Cauchy’s Theorem

Cauchy’s Theorem:

Let R be the closed region consisting of all points interior to and on the simple closed contour C.

If f is analytic in R and f' is continuous in R,

\int_C f(z)\,dz=0

(This is the precursor of Cauchy-Goursat Theorem, which allows us to drop the condition that f' is continuous.)

Proof Using Green’s Theorem:

Let C denote a positively oriented simple closed contour z=z(t), a\leq t\leq b.

\int_C f(z)\,d(z)=\int_a^b f[z(t)]z'(t)\,dt where f(z)=u(x,y)+iv(x,y) and z(t)=x(t)+iy(t). Thus

\begin{aligned} \int_C f(z)\,dz&=\int_a^b (u+iv)(x'+iy')\,dt\\    &=\int_a^b (ux'-vy')\,dt+i\int_a^b (vx'+uy')\,dt\\    &=\int_C u\,dx-v\,dy+i\int_C v\,dx+u\,dy    \end{aligned}

Next, we need Green’s Theorem:

\int_C P\,dx+Q\,dy=\iint_R (Q_x-P_y)\,dA

By assumption f' is continuous in R, thus the first-order partial derivatives of u and v are also continous. This is exactly what we need for Green’s Theorem.

Continuing from above, we get \int_C f(z)\,dz=\iint_R (-v_x-u_y)\,dA+i\iint_R(u_x-v_y)\,dA which is exactly zero in view of the Cauchy-Riemann equations u_x=v_y, u_y=-v_x!

Mapping of Infinite Vertical Strip by the Exponential Function

This post is continued from a previous post on how to map an open region between two circles to a vertical strip. We wish to map the infinite vertical strip 0<\text{Re}(z)<1 onto the upper half plane. Reference book is Complex Variables and Applications, Example 3 page 44.

First, we need to map the vertical strip onto the horizontal strip 0<y<\pi. This is easily accomplished by w=\pi iz. The factor i is responsible for the rotation (90 degree anticlockwise), while the factor \pi is responsible for the scaling.

Let us consider the exponential mapping w=e^z=e^{x+iy}=e^x\cdot e^{iy}. Consider line y=k. This line will be mapped onto the ray (from the origin) with argument k. Since 0<y<\pi, the image is a collection of rays of arguments 0<\theta<\pi that “sweeps” across and covers the entire upper half plane.

In conclusion, the map f(z)=e^{\pi i z} will do the job for mapping the vertical strip 0<x<1 onto the upper half plane y>0.

Image of Vertical Strip under Inversion

This is a slight generalisation of Example 3 in Churchill’s Complex Variables and Applications.

Consider the infinite vertical strip c_1<x<c_2, under the transformation w=1/z, where c_1, c_2 are of the same sign.

When x=c_1, note that by arguments similar to earlier analysis, we have x=\frac{u}{u^2+v^2}.

u^2+v^2=\frac{1}{c_1}u, upon completing the square, becomes

(u-\frac{1}{2c_1})^2+v^2=(\frac{1}{2c_1})^2 — Circle 1

Similarly the line x=c_2 is transformed into:

(u-\frac{1}{2c_2})^2+v^2=(\frac{1}{2c_2})^2 — Circle 2

Note that as x gets larger, the radius of the resultant circle gets smaller.

Thus, the resultant image is the (open) region between the two circles.

The converse holds too, i.e. the region between two circles will be mapped back to the vertical strip by inversion.


Find an analytic isomorphism from the open region between the two circles |z|=1 and |z-\frac{1}{2}|=\frac{1}{2} to the vertical strip 0<\text{Re}(z)<1.

First, we “shift” the circles to the left by 1 unit via the map w=z-1. Now we are in the situation of our above analysis, thus inversion w=1/z maps the region to the vertical strip -1<x<-1/2. The map w=-z (reflection), followed by w=z-\frac{1}{2}, finally finishing up with a scaling of factor 2 brings us to the vertical strip desired.

Composition of all the above functions leads us to the desired transformation f(z)=2(\frac{-1}{z-1}-\frac 12)=\frac{-1-z}{z-1} to the very nice strip 0<\text{Re}(x)<1.

We will mention an analytic isomorphism of this vertical strip to the upper half plane \text{Im} z>0 in a subsequent blog post.

Linear Fractional Transformation (Mobius Transformation)

The transformation w=\frac{az+b}{cz+d}, with ad-bc\neq 0, and a,b,c,d are complex constants, is called a linear fractional transformation, or Mobius transformation.

One key property of linear fractional transformations is that it transforms circles and lines into circles and lines.

Let us find the linear fractional transformation that maps the points z_1=2, z_2=i, z_3=-2 onto the points w_1=1, w_2=i, w_3=-1. (Question taken from Complex Variables and Applications (Brown and Churchill))

Solution: w=\frac{3z+2i}{iz+6}

What we have to do is basically solve the three simultaneous equations arising from w=\frac{az+b}{cz+d}, namely 1=\frac{2a+b}{2c+d}, i=\frac{ia+b}{ic+d} and -1=\frac{-2a+b}{-2c+d}.

Eventually we can have all the variables in terms of c: a=-3ic, b=2c, d=-6ic. Substituting back into the Mobius Transformation gives us the answer.