Isolated Singularities & Residue Basics

Solution:

This is a pole of order 2. Using the formula with $m=2$:

$\text{Res}(f,0) = \frac{1}{1!}\lim_{z\to 0} \frac{d}{dz}[z^2 \cdot \frac{e^z}{z^2}] = \lim_{z\to 0} \frac{d}{dz}[e^z] = \lim_{z\to 0} e^z = 1$

Solution:

Singularities: simple pole at $z=0$, pole of order 2 at $z=1$.

At $z=0$ (simple pole): $\text{Res}(f,0) = \lim_{z\to 0} z \cdot \frac{1}{z(z-1)^2} = \frac{1}{(0-1)^2} = 1$

At $z=1$ (pole of order 2): $\text{Res}(f,1) = \lim_{z\to 1} \frac{d}{dz}[(z-1)^2 \cdot \frac{1}{z(z-1)^2}] = \lim_{z\to 1} \frac{d}{dz}[\frac{1}{z}] = \lim_{z\to 1}[-\frac{1}{z^2}] = -1$

Solution:

Singularities: $z = \pm i$, both inside $|z|=2$.

Using the formula: $\text{Res}(f,i) = \frac{i}{2i} = \frac{1}{2}$, $\text{Res}(f,-i) = \frac{-i}{-2i} = \frac{1}{2}$

By residue theorem: $\oint_{|z|=2} \frac{z}{z^2+1}dz = 2\pi i(\frac{1}{2} + \frac{1}{2}) = 2\pi i$

Solution:

This is a pole of order 2. Using $\cos z = 1 - \frac{z^2}{2} + \cdots$:

$\frac{\cos z}{z^2} = \frac{1}{z^2} - \frac{1}{2} + \cdots$

The coefficient of $z^{-1}$ is 0, so $\text{Res}(f,0) = 0$.

Solution:

Using $e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!z^n} = 1 + \frac{1}{z} + \frac{1}{2z^2} + \cdots$

The Laurent series has infinitely many negative powers, so $z = 0$ is an essential singularity.

Solution:

This is a pole of order 3. Let $g(z) = (z-1)^3 f(z) = \frac{z^2 + 1}{z+2}$

Then $\text{Res}(f,1) = \frac{1}{2!}g''(1)$

$g'(z) = \frac{(z+2)(2z) - (z^2+1)}{(z+2)^2} = \frac{2z^2 + 4z - z^2 - 1}{(z+2)^2} = \frac{z^2 + 4z - 1}{(z+2)^2}$

$g''(z) = \frac{(z+2)^2(2z+4) - (z^2+4z-1)(2(z+2))}{(z+2)^4} = \frac{2(z+2)^3 - 2(z^2+4z-1)(z+2)}{(z+2)^4}$

At $z = 1$: $g''(1) = \frac{2 \cdot 27 - 2 \cdot 4 \cdot 3}{81} = \frac{54 - 24}{81} = \frac{30}{81} = \frac{10}{27}$

So $\text{Res}(f,1) = \frac{1}{2} \cdot \frac{10}{27} = \frac{5}{27}$

Solution:

Using $\sin w = \sum_{n=0}^{\infty} \frac{(-1)^n w^{2n+1}}{(2n+1)!}$:

$\sin(1/z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!z^{2n+1}} = \frac{1}{z} - \frac{1}{6z^3} + \frac{1}{120z^5} - \cdots$

This has infinitely many negative powers, so $z = 0$ is an essential singularity.

Solution:

The residue at infinity is defined as $\text{Res}(f,\infty) = -\text{Res}(f(1/z)/z^2, 0)$

$f(1/z) = \frac{1}{(1/z)(1/z - 1)} = \frac{z^2}{1-z}$

$\frac{f(1/z)}{z^2} = \frac{1}{1-z} = 1 + z + z^2 + \cdots$ for $|z| < 1$

The coefficient of $z^{-1}$ is 0, so $\text{Res}(f,\infty) = 0$

Solution:

Using $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$:

$\frac{e^z}{z^3} = \sum_{n=0}^{\infty} \frac{z^{n-3}}{n!} = \sum_{n=-3}^{\infty} \frac{z^n}{(n+3)!}$

The coefficient of $z^{-1}$ corresponds to $n = -1$: $\frac{1}{(-1+3)!} = \frac{1}{2!} = \frac{1}{2}$

Therefore, $\text{Res}(f,0) = \frac{1}{2}$.

Solution:

Singularities: $z = 0$ and $z = k\pi$ for integers $k \neq 0$ (zeros of $\sin z$).

At $z = k\pi$ ($k \neq 0$), $\cot z$ has simple poles, so $f$ has simple poles with residues computed using L'Hôpital's rule.

At $z = 0$, expand $\cot z = \frac{\cos z}{\sin z} = \frac{1}{z} - \frac{z}{3} - \cdots$, so $f(z) = \frac{1}{z^3} - \frac{1}{3z} - \cdots$

Therefore, $\text{Res}(f,0) = -\frac{1}{3}$.

Solution:

If $f$ has a pole of order $m$, then $f(z) = \frac{g(z)}{(z-z_0)^m}$ where $g(z_0) \neq 0$ and $g$ is analytic.

Then $1/f(z) = \frac{(z-z_0)^m}{g(z)}$, and since $g(z_0) \neq 0$, $1/g$ is analytic at $z_0$.

Therefore $1/f$ has a zero of order $m$ at $z_0$.

Solution:

$f(z) = z^2(z - \frac{z^3}{6} + \cdots) = z^3 - \frac{z^5}{6} + \cdots$

The first non-zero term is $z^3$, so $z = 0$ is a zero of order 3.

Solution:

This is a pole of order 2. Let $g(z) = (z-1)^2 f(z) = \frac{z^2}{z+1}$

$g'(z) = \frac{(z+1)(2z) - z^2}{(z+1)^2} = \frac{2z^2 + 2z - z^2}{(z+1)^2} = \frac{z^2 + 2z}{(z+1)^2}$

$g'(1) = \frac{1 + 2}{4} = \frac{3}{4}$

Therefore: $\text{Res}(f,1) = \frac{1}{1!} \cdot \frac{3}{4} = \frac{3}{4}$

Solution:

Using $\sin z = z - \frac{z^3}{6} + \frac{z^5}{120} - \cdots$:

$f(z) = \frac{z - z^3/6 + z^5/120 - \cdots - z}{z^3} = -\frac{1}{6} + \frac{z^2}{120} - \cdots$

The Laurent series has no negative powers, so $z = 0$ is a removable singularity.

Solution:

Singularities: $z = 0$ and $z = k\pi$ for integers $k \neq 0$.

At $z = k\pi$ ($k \neq 0$), simple poles with residues computed using L'Hôpital's rule.

At $z = 0$, expand $\sin z = z - z^3/6 + \cdots$, so $f(z) = \frac{1}{z^2} - \frac{1}{6} + \cdots$, showing a pole of order 2.

Solution:

Using $\cos z = 1 - \frac{z^2}{2} + \frac{z^4}{24} - \cdots$:

$f(z) = \frac{z^2/2 - z^4/24 + \cdots}{z^2} = \frac{1}{2} - \frac{z^2}{24} + \cdots$

No negative powers, so $z = 0$ is a removable singularity.

Solution:

Using $e^z = 1 + z + \frac{z^2}{2} + \cdots$:

$f(z) = \frac{z + z^2/2 + z^3/6 + \cdots}{z^2} = \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \cdots$

Therefore, $\text{Res}(f,0) = 1$ (coefficient of $z^{-1}$).

Solution:

If $f$ has a pole of order $m$, then $f(z) = \frac{g(z)}{(z-z_0)^m}$ where $g(z_0) \neq 0$ and $g$ is analytic.

Then $(z-z_0)^m f(z) = g(z)$, which is analytic at $z_0$ and $g(z_0) \neq 0$.

Solution:

Singularities: $z = 0$ (pole of order 3) and $z = k\pi$ for $k \neq 0$ (simple poles).

At $z = k\pi$, the residue is $(-1)^k/(k\pi)^2$.

At $z = 0$, expand to find the Laurent series and extract the residue.

Solution:

Using $e^z = 1 + z + z^2/2 + \cdots$:

$f(z) = \frac{-z - z^2/2 - \cdots}{z} = -1 - z/2 - \cdots$

No negative powers, so $z = 0$ is a removable singularity.

Solution:

Using $\sin z = z - z^3/6 + z^5/120 - \cdots$:

$f(z) = \frac{1}{z^3} - \frac{1}{6z} + \frac{z}{120} - \cdots$

The coefficient of $z^{-1}$ is $-1/6$, so $\text{Res}(f,0) = -1/6$.

Solution:

Singularities: $z = 0$ (pole of order 3) and $z = 1$ (simple pole).

At $z = 1$: $\text{Res}(f,1) = \lim_{z\to 1} (z-1)f(z) = \frac{2}{1} = 2$

At $z = 0$: Use the derivative formula to compute the residue at the pole of order 3.

Solution:

This is a pole of order 3. Let $g(z) = (z-1)^3 f(z) = e^z$

$\text{Res}(f,1) = \frac{1}{2!}g''(1) = \frac{1}{2} \cdot e^1 = \frac{e}{2}$

Solution:

By Casorati-Weierstrass theorem, $f$ takes values arbitrarily close to any complex number near $z_0$.

If the limit existed, $f$ would be bounded near $z_0$, contradicting the essential singularity behavior.

Solution:

Using $\cos z = 1 - z^2/2 + z^4/24 - \cdots$:

$f(z) = \frac{z^2/2 - z^4/24 + \cdots}{z^3} = \frac{1}{2z} - \frac{z}{24} + \cdots$

Therefore, $\text{Res}(f,0) = 1/2$.

Solution:

If $z_0$ is a removable singularity, then the Laurent series has no negative powers, so $f$ can be defined at $z_0$ to be analytic there.

By continuity of analytic functions, the limit exists and equals $f(z_0)$.

Solution:

Singularities: $z = \pm i$ (each is a pole of order 2).

At $z = i$: $\text{Res}(f,i) = \lim_{z\to i} \frac{d}{dz}\left[(z-i)^2 f(z)\right]$

Computing gives $\text{Res}(f,i) = -i/4$ and $\text{Res}(f,-i) = i/4$.

Solution:

The function $f(z) = e^{1/z}$ has an essential singularity at $z = 0$ (its Laurent series has infinitely many negative powers).

By Casorati-Weierstrass theorem, for any $w \in \mathbb{C}$ and any $\varepsilon > 0$, there exists $z$ with $0 < |z| < \varepsilon$ such that $|f(z) - w| < \delta$ for any $\delta > 0$.

To see this explicitly, consider sequences approaching 0 from different directions:

For $z_n = 1/(2\pi in)$: $f(z_n) = e^{2\pi in} = 1$

For $z_n = 1/(2\pi in + \pi i)$: $f(z_n) = e^{2\pi in + \pi i} = -1$

By choosing appropriate sequences, we can make $f(z)$ approach any complex number.

Solution:

Suppose $|f(z)| \leq M$ for all $z$ in some deleted neighborhood of $z_0$.

Consider $g(z) = 1/(f(z) - w)$ for some $w$. If $f$ is bounded away from $w$, then $g$ would be bounded.

However, by Casorati-Weierstrass, $f$ takes values arbitrarily close to any $w$, so $g$ cannot be bounded.

This contradiction shows that $f$ cannot be bounded in any deleted neighborhood of an essential singularity.

This is a stronger statement than Casorati-Weierstrass: not only does $f$ take values dense in $\mathbb{C}$, but it's also unbounded.