Exponents

Exponential function

We sometimes write

\exp(z) = e^z = \sum_{n=0}^\infty \frac{z^n}{n!}

to show that e^z is a function of z. This series converges everywhere (by the ratio test).

We can prove Eulers formula.

Theorem

Let exp(z) be the exponential function. Then

• \frac{d}{dz}\exp(z) = \exp(z)

• \exp(z_1 + z_2) = \exp(z_1)\exp(z_2).

Say we want to find the inverse function of \exp. We have the problem that \exp is not injetive. In fact, for every w in \exp(z) = w, there are infinetly many z to make it true.

Complex Logarithm

We will call the inverse function \log. We say

\log(z) = w \iff z = \exp(w).

If we write z = re^{i\theta} and w = x +yi, then we see

\log(z) = w \iff re^{i\theta} = e^{x + yi}.

Thus r = e^x and y \in \arg(z). So x = \ln r, and y \in \arg(z). So

\log(z) = \ln z + i\arg(z).

Definition

The principle branch of the logarithm is given by

\text{Log}(z) = \ln |z| + i\arg(z).

Theorem

• \log(z_1z_2) = \log(z_1) + \log(z_2)

• \log(1/z) = -\log(z)

Proof

(b) Recall \arg(1/z) = -\arg(z). Then we have

\begin{align*} \log(1/z) &= \ln(|1/z|) + \arg(1/z)i\\ &= \end{align*}

Things aren’t always as easy as you want them to be. - Professor Griffen.

The principle value \text{Log}(z) does not have these properties as it is bounded. But when does it?

Theorem

• \text{Log}(z_1z_2) = \text{Log}(z_1) + \text{Log}(z_2) when

-\pi < \text{Arg}(z_1) + \text{Arg}(z_2) \leq \pi

• \log(1/z) = -\log(z)

We can create other branches of \log with \log_\alpha(z) in the strip \alpha < \theta \leq \alpha + 2\pi.

So \log_{-\pi}(z) = \text{Log}(z). To calculate another branch

Example

Complex exponents

We can give meaning to statements like i^{2+3i}.

The principle branch of z^c is

f(z) = \exp(c\text{Log}(z)).

Example

Take for example (1+i)^i, we get

\ln(\sqrt{2}) + i(\pi/4 + 2\pi n).

which gives

e^{-\pi/4 + 2\pi n}(\cos\ln\sqrt{2} + i\sin\ln\sqrt{2})

with a principle value e^{-\pi/4}(\cos\ln\sqrt{2} + i\sin\ln\sqrt{2}).

Theorem

These properties hold.

• z^{-c} = \frac{1}{z^c}

• \frac{z^c}{z^d} = z^{c-d}

• z^cz^d = z^{c+d}

• (z^c)^n = z^{nc}

Proof

Proofs for each statement

---

(a) Note that \exp(-w) = \frac{1}{\exp(w)}. Then

z^{-c} = \exp((-c)\log)

---

(b) Note since \exp(w+z) = \exp(w)\exp(z), we have

\begin{align*} z^{c+d} &= \exp((c+d)\log{z})\\ &= \exp(c\log{z}) + \exp(d\log{z})\\ &= z^c + z^d. \end{align*}

---

(c) Since \exp(w)^n = \exp(nw) for n being an integer, we have

(z^{c})^n = \exp^n((c)\log{z})\\ = \exp(nc\log(z))\\ = z^{cn}

Warning

For complex numbers, (z^w)^v \not = (z^{wv}) for some values.

(i^2)^i = \exp(i\log(-1)) = \exp(i(i(\pi + 2\pi n)))

But

i^{2i} = \exp(-(\pi + 4\pi n))

Trig and hyperbolic functions

Define

Definition

The complex sine and cosine functions

\sin(z) = \sum^\infty_{n=0} \frac{(-1)^nz^{2n+1}}{(2n+1)!}

\cos(z) = \sum^\infty_{n=0} \frac{(-1)^nz^{2n}}{(2n)!}

These functions are both entire. [TO-DO] proof. Do the same derivative rules apply?

Theorem

The derivative rules hold. That is

• \frac{d}{dz}\cos(z) = -\sin (z)

• \frac{d}{dz} \sin(z) = \cos(z)

Properties

Theorem

• \cos(-z) = \cos(z)

• \sin(-z) = - \sin(z)

Theorem

We have e^{iz} = \cos z + i \sin z.

Theorem

We have

\cos z = \frac{e^{iz} + e^{-iz}}{2}

\sin z = \frac{e^{iz} - e^{-iz}}{2}

Complex integration

Let f(z) be a complex function, then there exists a f(z) = u(x) + iv(x). So

Definition

For a complex function f(z) we have

\int f(z) = \int u(x) + i \int v(x)

Example

For f(z) = (t + i)^2 we have

\begin{align*} \int_0^1 (t - i)^2 dt &= \left[\frac{1}{3}(t+i)^3\right] \end{align*}

Example

For f(z) = e^t\cos(t), let u = e^t and v = \cos(t). Then

\begin{align*} \int_0^{\pi/2} e^t\cos(t)\, dt &= \frac{e^{\pi/2 - 1}}{2} \end{align*}

Recall e^{it} = \cos t + i \sin t. Then

e^{t+it} = e^t\cos t + i e^t \sin t

We use eulers method and then separate into both real and complex parts.

Recall from lecture 7 that a parameterized curve in the complex plane is

C \colon z(t) = x(t) + i y(t)

for some continuous real valued functions. We say z(t) is differentiable if both x(t) and y(t) are.

We will integrate complex functions over contours, unions of smooth curves.

In order to do Riemann sums, we split our contour into many different intervals and then take a test under each contour.

\sum_{i=1}^n f(c_i)(z_i - z_{i-1}) = \sum_{i=1}f(c_i)\Delta z

Then we can define the integral.

\int_C f(z) dz = \lim_{n \to \infty} \sum_{i=1}f(c_i)\Delta z

Then we can do some coolio

\begin{align*} \int_C f(z) dz &= \lim_{n \to \infty} \sum_{i=1}f(c_i)\Delta z\\ &= \int_a^b f(z(t))z^\prime(t) dt \end{align*}

Example

For f(z) = z^2, and a contour line from 0 to 3i we need to intgerate. Here we choose the parameterization z(t) = 3ti for 0 \leq t \leq 1. This is clearly a smooth contour. Thus

\begin{align*} \int_C f(z) dz &= \int_0^1 (z(t))^2z^\prime(t) dt\\ &= \int_0^1 (3ti)^2(3i)\\ &= -9i \end{align*}

If we instead parametrize z(t) = t^2i where 0 \leq t \leq \sqrt{3}. Then z^\prime(t) = 2ti.

\begin{align*} \int_C f(z) dz &= \int_0^1 (z(t))^2z^\prime(t) dt\\ &= \int_0^{\sqrt{3}} (t^2i)^2(2ti)\\ &= -9i \end{align*}

Images of sin and cos

Recall

\cos z = \frac{e^{iz} + e^{-iz}}{2}

\sin z = \frac{e^{iz} - e^{-iz}}{2}

We also have

\cosh z = \frac{e^{z} + e^{-z}}{2}

\sinh z = \frac{e^{z} - e^{-z}}{2}

Theorem

We have the identity

\sin(x + yi) = \sin(x)\cosh(y) + i \cos(x)\sinh(y).

Proof

Example

Find the image of \{z \colon \Re (z) = 2\} under \sin z.

Let w = \sin(z) and z = 2 + yi. Then

u + vi = \sin(2 + yi).

Using our previous theorem

\begin{align*} u + vi &= \sin(2 + yi)\\ &= \sin(2)\cosh(y) + i\cos(2)\sinh(y) \end{align*}

Theorem (Deformation of contours)

Let C_1,C_2 be two simple closed positively oriented contours such that

• C_1 is in the interioir of C_2, and

• f(z) is analytic on some domain D that contains C_1 and C_2 and the area inbetween.

Then \int_{C_1} f(z) = \int_{C_2} f(z).

Proof

C_2 = A +B, and C_1 = E + F. Then

\int_{C_2} - \int_{C_1} = \int_A f + \int_B f - \int_E f - \int_F f

Corollary

Let C be a simple closed positively oriented contour containing some point z_0 \in \mathbb{C} in its interioir. Then

• \displaystyle\int_C \frac{1}{z - z_0} dz = 2\pi i

• \displaystyle\int_C (z - z_0)^n dz = 0 for all integers n \not = -1.

Proof

The function f(z) = 1/(z - z_0) is always analytic except for z = z_0, so we can deform the contour to some circle containing z_0.

\int_C \frac{1}{z - z_0} dz = \int_{C^+_r(z_0)}\frac{1}{z - z_0} dz = 2\pi i.

The function f(z) = (z- z_0)^n is entire for every nonnegative integer n, so it is always 0. When n is negative and not 1, we choose a paramitization and obtain 0.

Theorem (Extended Cauchy-Goursat)

Let C and C_1,\dots,C_k be simple closed positively oriented contours such that

• C_i are all in the interioir of C

• The interiors of C_i are all disjoint.

• f(z) is analytic on some domain D that contains all these contours and the region between C and C_i.

Then \int_{C_1} f(z) = \sum_{i=1}^k \int_{C_i} f(z).

Fundamental Theorems of Integration

So far we have only done definite integrals over contours. Lets expand our purview.

Theorem

If f(z) is an analytic function in a simply connected domain D, such that z_0 is in D and C is any contour in D with an initial point z_0 and terminal point z, then

• \displaystyle F(z) = \int_C f(w) dw

• F(z) is a well defined function of z which is analytic on D such that F^\prime(z) = f(z).

We will often write F(z) = \int^z_{z_0} f(w) dw to mean that we are free to choose any contour C. This integral is called the antiderivative.

Proof

We first need to show that using any two contours will lead us to the same integral.

Then we must prove that the derivative is equivalent to the function we began with.

F(z + \Delta z) - F(z) = \int_{C_1} f(w) dw - \int_{C_2} f(w) dw

Due to continuity, we can keep shrinking \Delta z so that |\Delta z| > \delta. Since |w - z| > \delta, we have that |f(w) - f(z)| < \varepsilon.

So F^\prime(z) = f(z).

If C is a positively oriented curve, then integrating over |dz| gives the length of the curve C.

\int_C |dz| = \text{len } C

Bounding integrals

Theorem (Bounding integrals)

If f(z) is continuous on a contour C with a maximum of M.

\begin{align*} \left|\int_C f(z) dz\right| &\leq \int_C |f(z)| |dz|\\ &\leq M\int_C |dz|\\ &= M \cdot \text{len } C \end{align*}

Theorem (Definite integrals)

Let f be analytic in a simply connected domain D. If z_1 and z_2 are points in D joined by a contour C, then

\int_C f(z) dz = \int_{z_1}^{z_2} f(z) dz

Cuachys integral formula

Theorem (Cauchy integral formula)

Let f be analytic in a simply connected domain D, and let C be a simply closed positively oriented contour. If z_0 is in the interior of C,

f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_0} dz.

Proof

Since z_0 is in the interior of your contour, there exists some r > 0 such that C_r(z_0) is contained in the interior of C. Then

\begin{align*} \frac{1}{2\pi i} \int_{C_{r}^+(z_0)} \frac{f(z)}{z - z_0} dz &=\\ &= \frac{f(z_0)}{2\pi i} \int_{C_{r}^+(z_0)} \frac{1}{z - z_0}dz\\ &= \frac{f(z_0)}{2\pi i} 2\pi i = f(z_0) \end{align*}

It comes down to proving

\int_{C_{r}^+(z_0)} \frac{f(z)}{z - z_0} dz = \int_{C_{r}^+(z_0)} \frac{f(z_0)}{z - z_0} dz

Since f(z) is continuous in D, for every \varepsilon > 0 there exists a \delta > 0 such that

|z - z_0| < \delta \implies |f(z) - f(z_0)| < \varepsilon.

Then

\begin{align*} \left| \int_{C_{\delta/2}^+(z_0)} \frac{f(z)}{z - z_0} dz - \int_{C_{\delta/2}^+(z_0)} \frac{f(z_0)}{z - z_0} dz\right| &=\\ &= \left| \int_{C_{\delta/2}^+(z_0)} \frac{f(z) - f(z_0)}{z - z_0} dz \right|\\ &\leq \int_{C_{\delta/2}^+(z_0)} \frac{|f(z) - f(z_0)|}{|z - z_0|} dz\\ &\leq \frac{\varepsilon}{\delta/2} \int_{C_{\delta/2}^+(z_0)} |dz|\\ &\leq \frac{\varepsilon}{\delta/2}(2\pi (\delta/2)) = 2\pi\varepsilon \end{align*}

Then the difference of the two integrals is 0, and thus

\int_{C_{r}^+(z_0)} \frac{f(z)}{z - z_0} dz = \int_{C_{r}^+(z_0)} \frac{f(z_0)}{z - z_0} dz.

Divide each side by 2\pi i and use deformation of contours to prove

f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_0} dz.

Example

Find

\int_{C_2^+(0)} \frac{e^z}{z - i} dz.

Here z_0 = i and f(z) = e^z. Then by the Cauchy integral theorem

\int_{C_2^+(0)} \frac{e^z}{z - i} dz = 2\pi e^i.

Theorem (Cauchy integral formula for nth derivatives)

Let f be analytic in a simply connected domain D, and let C be a simply closed positively oriented contour. If z_0 is in the interior of C, then for any nonnegative integer n \geq 0,

f^(n)(z_0) = \frac{n!}{2\pi i} \int_C \frac{f(z)}{(z - z_0)^{n+1}} dz.

Review

When is a function analytic?

If f is analytic, then the CR equations hold

u_x = v_Y

u_y = - v_x

A function must be continuous and differentiable.

[Study definitions]

Definition

Let z,w \in \mathbb{C}, and f \colon \mathbb{C} \to \mathbb{C} be a function. The limit of f at z is w if for all D_\varepsilon(w), there exists D_\delta(z) such that f(D_\delta(z)) \subset D_\varepsilon(w).

Equivalently, if |z - z_0| < \delta then |f(z) - w_0| < \varepsilon.

Definition

A complex function is continuous if for all z \in D, we have

\lim_{z \to z_0} = f(z_0).

Definition

Let z_0 \in \mathbb{C}, and f be a complex valued function. Then the derivative of f is defined

f^\prime(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}

if limit exists. If f^\prime(z_0) exists, we say f is differentiable at z_0.

Definition

Let z_0 \in \mathbb{C} a7nd f be a complex valued function. The function f is holomorphic at a point z_0 if it is differentiable at at every point in some neighborhood D_\varepsilon(z_0).

A function is harmonic if u_{xx} + u_{yy} = 0.

If f(z) = u + vi, then both u and v are harmonic.

Given a harmonic function, you can find a harmonic conjugate.

Mareras and Liouville’s Theorem

Theorem (Marera’s Theorem)

Let f(z) be a continuous function on a simply connected domain D, and let

\int_C f(z) dz = 0

for all simply closed contours C. Then f(z) is an analytic function.

The idea is that we prove f is analytic starting from a weaker starting condition.

Theorem (Gauss’s Mean Value Theorem)

Let f(z) be an analytic function on a simply connected domain D containing a circle C_r(z_0), then

f(z_0) = \frac{1}{2\pi} \int_0^2\pi f(z_0 + re^{it}) dt.

The idea is that the value of the function at the center is the ‘average’ of the function across a circle around z_0.

Proof

The Cauchy integral formula implies

f(z) = \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{f(z_0)}{z - z_0}.

Then we have

\begin{align*} f(z) &= \frac{1}{2\pi i} \int_{C_r(z_0)} \frac{f(z_0)}{z - z_0}\\ &= \frac{1}{2\pi i} \int^{2\pi}_0 \frac{f(z_0 + re^{it})}{z_0 + re^{it} - z_0} ire^{it} dt\\ &= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{it}) dt \end{align*}

Lemma (Maximum Modulus Principle)

If f(x,y) is continuous on a bounded domain D, then f obtains its maximum value of either at a point m \in D, or on the boundary of D.

Theorem (Local Maximum Modulus Principle)

Let f(z) be analytic on some closed disk D_R(z_0). If f(z) is non-constant, then the maximum of the function is not obtained on the disk.

Proof (Local Maximum Modulus Principle)

Assume \max|f(z)| = |f(z_0)|. Then |f(z)| \leq |f(z_0)| for all z = D_R(z_0). We will show this forces f(z) to be constant.

By Gauss’s Mean Value Theorem, for 0 < r < R, we have

\begin{align*} |f(z_0)| &= \left| \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{it}) dt \right| \\ &\leq \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{it} dt)\\ &\leq \frac{1}{2\pi} \int_0^{2\pi} |f(z_0)| dt\\ &= |f(z_0)| \end{align*}

Then clearly all of our inequalities are equivalent. Thus

\frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{it}) dt = \frac{1}{2\pi} \int_0^{2\pi} |f(z_0)| dt

Since these integrals must be equivalent, we can subtract to obtain

\int_0^{2\pi} \left| f(z_0) - f(z_0 + re^{it}) \right| = 0.

Thus f(z_0) = f(z_0 + re^{it}) for all t. Then the function must be a constant.

Theorem (Maximum Modulus Principle)

Let f(z) be analytic and non-constant on some closed disk D_R(z_0). Then \max |f(z)| is not obtained in D.

Proof (Maximum Modulus Principle)

Assume \max|f(z)| = |f(z_0)| where z_0 = D. Let P be some path from z to z_0. Then P can be covered by finitely many disks D_0,D_1,\dots,D_m containing z_0,z_1,\dots,z_m, such that for all i \leq m

z_i \in D_i \cap D_{i-1}.

By our Local Maximum Modulus Principle, we have that f(z) is constant on D_0. Thus f(z_0) = f(z_1). Since f(z_0) is the maximum, we have that f(z_1) is the maximum of D_1. Since the function obtains a maximum on D_1, we have that the function is constant on D_1. We apply this argument m times to obtain that f(z_0) = f(z).

Since z was arbitrary, we have that f(z) is constant, and thus a contradiction.

Now an example.

Example (Maximum Modulus Principle)

Find the maximum of |az + b| on D_r(0) for a,b \in \mathbb{C} fixed. Let f(z) = az + b.

If a = 0 then clearly f(z) = b, and the maximum of f(z) is b.

If a \not = 0, and b = 0 then f is not constant on D_r(0). Then

|f(z)| = |za| = |a| \cdot |z| \leq |z| \cdot r.

If a \not = 0, and b \not = 0 then the maximum is obtained at the boundary. Let z = re^{it} for 0 \leq t \leq 2pi.

Then the maximum |f(z(t))| = |a| \cdot r + |b|. Is this obtainable?

Theorem (Cauchys Inequality)

Let f(z) be analytic in a simply connected domain D containing the a circle C_R(z_0). If |f(z)| \leq M for all z \in C_R(z_0), then

\left|f^{(n)}(z_0)\right| \leq \frac{n!M}{R^n}

for all 0 \leq n.

Proof (Cauchys Inequality)

We have from Cauchy’s integral formula,

\begin{align*} f^{(n)}(z_0) &= \frac{n!}{2\pi i} \int_{C^+_R(z_0)} \frac{f(z)}{(z - z_0)^{n+1}} dz\\ |f^{(n)}(z_0)| &= \frac{n!}{2\pi i} \left| \int_{C^+_R(z_0)} \frac{f(z)}{(z - z_0)^{n+1}} dz \right|\\ &\leq \frac{n!}{2\pi} \int_{C^+_R(z_0)} \frac{|f(z)|}{|(z - z_0)^{n+1}|} dz\\ &\leq \frac{n!}{2\pi} \int_{C^+_R(z_0)} \frac{M}{R^{n+1}} dz\\ &= \frac{n!M}{R^n}. \end{align*}

Thus

\left|f^{(n)}(z_0)\right| \leq \frac{n!M}{R^n}.

Theorem (Liouville’s Theorem)

Let f(z) be entire and |f(z)| \leq M for all z. Then f(z) is a constant function.

Proof (Liouville’s Theorem)

Use Cauchy’s inequality for n = 1. Since \mathbb{C} is a simply connected domain, we have

|f^\prime(z)| \leq \frac{1!M}{R^1} = \frac{M}{R}

for arbitrary R. Thus |f^\prime(z)| = 0, and thus f^\prime(z) = 0, which implies f is a constant function.

Theorem (Fundamental Theorem of Algebra)

If P(z) is a polynomial function of degree n \geq 1, then their exists z_0 such that P(z_0) = 0.

Proof (Fundamental Theorem of Algebra)

Assume P(z) has no root. Let f(z) = 1/P(z). Since P(z) is entire and never zero, we have that f(z) is entire and nonconstant.

Contradiction.

Theorem (Little Picard’s Theorem)

If f(z) is entire with an image notcontaining at least two points, then f(z) is constant.

Power series

Definition

If f(z) is analytic at a point z = a, then the Taylor Series of f centered at \alpha is

\sum

Theorem (Taylors Theorem)

Suppose f(z) is analytic on a domain D and D_r(\alpha) is a disk contained in D. Then the Taylor series for f centered at \alpha converges to f(z).

Convergence and divergence on disks.

Example

\frac{1}{z -2} = \frac{(1)}{(z-3) + 1} = \frac{1}{1 -(z - 3)}

Laurent series

If a function f(z) is not analytic at a point \alpha, then it wont have a taylor series. However, we may have a trick for this

• e^z = \sum\frac{1}{n!}z^n

Definition

A Laurent series is a convergent series

\sum_{n=-\infty}^\infty c_n(z-a)^n

where c_n are complex numbers. This may be formally strange, so we express it as

\sum_{n=1}^\infty c_{-n}(z-a)^{-n} + c_0 + \sum_{n=1}^\infty c_n(z-a)^n

Definition

An annulus centered at z = \alpha is

A(\alpha,r,R) = \{z \mid r < |z-a| < R\}

Theorem (Laurent’s Theorem)

Suppose f(z) is analytic on an annulus A = A(\alpha,r,R). For all z \in A, f(z) has a Laurent series expansion

f(z) = \sum_{n=-\infty}^\infty c_n(z-a)^n

where for all r < \rho < R,

c_n = \frac{1}{2\pi i} \int_{C_\rho^+(\alpha)} \frac{}{}dz

Theorem (Uniqueness of Laurent’s Theorem)

Suppose f(z) is analytic on an annulus A = A(\alpha,r,R) with Laurent series $$. Then

• If f(z) = \sum_{n=-\infty}^\infty b_n(z-a)^n then b_n = c_n for all n \in \mathbb{N}.

• The derivative f^\prime(z) = \sum_{n=-\infty}^\infty nc_n(z-\alpha)^{n-1} for all z \in A(\alpha,r,R).

Example

Find the Laurent series for \exp(-1/z^2) centered at \alpha = 0

Remeber

\frac{1}{1-z} = -\sum_{n=1}^\infty z^{-n}.

Poles and Zeros

Corollary

If f(z) has a zero of order m at \alpha and g(z) has a zero of order n at \alpha, then f(z)g(z) has a zero of order m+n at \alpha.

Corollary

If f(z) has a pole of order m at \alpha and g(z) has a zero of order n at \alpha, then

• If m > n then f(z)g(z) has a pole of order m-n at \alpha.

• If m < n then f(z)g(z) has a zero of order n-m at \alpha.

• If m = n then f(z)g(z) has a removable singularity at \alpha.

Residue Theorem

Definition

If f(z) has a nonremovable singularity at z_0, then the coefficient a_{-1} of (z-z_0)^{-1} in the Laurent series is called the residue of f at z_0.

Remember how to do Laurent series.

Definition

A Laurent series is a convergent series

\sum_{n=-\infty}^\infty c_n(z-a)^n

where c_n are complex numbers. This may be formally strange, so we express it as

\sum_{n=1}^\infty c_{-n}(z-a)^{-n} + c_0 + \sum_{n=1}^\infty c_n(z-a)^n

Theorem

Let C be a simple closed contour inside a simply conn3ected domain D.

\int_C f(z) dz = 2 \pi i \sum_{i=1}^n \text{Res}[f,z_0]

Theorem

If f has a simple pole at z_0, then

\text{Res}[f,z_0] = \lim_{z\to z_0}(z - z_0)f(z)

Theorem

If f has a pole of order 2 at z_0, then

\text{Res}[f,z_0] = \lim_{z\to z_0} \frac{d}{dz} (z - z_0)^2f(z)

Proof

If f has a pole of order 2 at z_0, then

\text{Res}[f,z_0] = \lim_{z\to z_0} \frac{d}{dz} (z - z_0)^2f(z)

Theorem

If f has a pole of order 2 at z_0, then

\text{Res}[f,z_0] = \frac{1}{(n-1)!}\lim_{z\to z_0} \frac{d^{n-1}}{dz^{n-1}} (z - z_0)^nf(z)

Trig with residues

Given

\int_0^{2\pi} F(\cos t, \sin t)\,dt