Week 8 Worksheet - Solutions

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Laurent Series

1. 1. $\displaystyle \dfrac{1}{z-3} = \dfrac{1}{z} \dfrac{1}{1 - \frac{3}{z}} = \dfrac{1}{z} \sum_{n=0}^\infty \left( \frac{3}{z} \right)^n = \dfrac{1}{z} + \dfrac{3}{z^2} + \dfrac{9}{z^3} + \cdots$ whenever $\left| \dfrac{3}{z} \right| < 1$ i.e. $|z| > 3$.
   2. $\displaystyle \dfrac{1}{z(1-z)} = \dfrac{1}{z} \sum_{n=0}^\infty z^n = \dfrac{1}{z} + 1 + z + z^2 + \cdots$ whenever $|z| \l 1$.
   3. $\displaystyle z^3 \exp(1/z) = z^3 \sum_{n=0}^\infty \dfrac{1}{n! z^n} = z^3 + z^2 + \dfrac{z}{2!} + \dfrac{1}{3!} + \dfrac{1}{4! z} + \cdots$ on $\C \setminus \{0\}$. \item $\displaystyle \cos(1/z) = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)! z^{2n}} = 1 - \dfrac{1}{2! z^2} + \dfrac{1}{4! z^4} - \dfrac{1}{6! z^6} + \cdots$ on $\C \setminus \{0\}$.

2. First note that \[ \dfrac{1}{z+1} = \begin{cases} \displaystyle\dfrac{1}{1 - (-z)} = \sum_{n=0}^\infty (-z)^n & |z| \l 1 \\ \dfrac{1}{z} \dfrac{1}{1 - (-\frac{1}{z})} = \dfrac{1}{z} \displaystyle\sum_{n=0}^\infty \left( -\dfrac{1}{z} \right)^n & |z| > 1 \end{cases} \] and that \[ \dfrac{1}{z-3} = \begin{cases} -\dfrac{1}{3}\dfrac{1}{1 - \frac{z}{3}} = \displaystyle -\dfrac{1}{3} \sum_{n=0}^\infty \left( \dfrac{z}{3} \right)^n & |z| \l 3 \\ \dfrac{1}{z} \dfrac{1}{1 - \frac{3}{z}} = \dfrac{1}{z} \displaystyle\sum_{n=0}^\infty \left( \dfrac{3}{z} \right)^n & |z| > 3 \end{cases} \] using geometric series.

   $\displaystyle \sum_{n=0}^\infty (-z)^n - \dfrac{1}{3} \sum_{n=0}^\infty \left( \dfrac{z}{3} \right)^n$

   $\displaystyle \dfrac{1}{z} \sum_{n=0}^\infty \left( -\dfrac{1}{z} \right)^n - \dfrac{1}{3} \sum_{n=0}^\infty \left( \dfrac{z}{3} \right)^n$

   $\displaystyle \dfrac{1}{z} \sum_{n=0}^\infty \left( -\dfrac{1}{z} \right)^n + \dfrac{1}{z} \displaystyle\sum_{n=0}^\infty \left( \dfrac{3}{z} \right)^n$

3. 1. $\displaystyle \dfrac{1}{z^2(z-1)} = -\dfrac{1}{z^2} \dfrac{1}{1-z} = -\dfrac{1}{z^2} \sum_{n=0}^\infty z^n = \dfrac{1}{z^2} + \dfrac{1}{z} + 1 + z + z^2 + z^3 + \cdots$ on $\Ann(0,0,1)$
   2. First note that \[ \dfrac{1}{z} = \dfrac{1}{1 - (1-z)} = \sum_{n=0}^\infty (1-z)^n \] whenever $0 \l |z-1| \l 1$. Differentiating gives \[ - \dfrac{1}{z^2} = -\sum_{n=0}^\infty n(1-z)^{n-1} = -\sum_{n=0}^\infty (n+1) (1-z)^n \] whenever $0 \l |z-1| \l 1$. Then calculate that \begin{align*} \dfrac{1}{z^2(z-1)} = \dfrac{-1}{z-1} \left( - \dfrac{1}{z^2} \right) & = \dfrac{1}{z-1} \sum_{n=0}^\infty (n+1) (-1)^n (z-1)^n \\ &= \dfrac{1}{z-1} - 2 + 3(z-1) - 4(z-1)^2 + \cdots \end{align*} for $0 \l |z-1| \l 1$.
4. 1. Just $\dfrac{1}{(z-1)^2}$
   2. $\dfrac{1}{z-1} = -\displaystyle \sum_{n=0}^\infty z^n$ for $|z| \l 1$ and so \[ \frac{1}{(z-1)^2} = - \left( \dfrac{1}{z-1} \right)' = -\sum_{n=0}^\infty n z^{n-1} = - \sum_{n=0}^\infty (n+1)z^n \] on $\Ann(0,0,1)$. (Note that $f$ is holomorphic on $\ball(0,1)$.)
   3. From part b) we have \[ - \sum_{n=0}^\infty \dfrac{n+1}{z^n} = \dfrac{1}{(\frac{1}{z} - 1)^2} = z^2 \dfrac{1}{(1-z)^2} \] whenever $|z| > 1$. Thus \[ \dfrac{1}{(1-z)^2} = \dfrac{1}{z^2} \sum_{n=0}^\infty \dfrac{n+1}{z^n} \] on $\Ann(0,1,\infty)$.

Isolated Singularities

5. 1. $z^2 + 1 = (z+i)(z-i)$ has simple zeroes at $i$ and $-i$ so $f$ has simple poles at $i$ and $-i$.
   2. $z^4 + 16 = (z - 2e^{i \pi/4})(z - 2e^{3i \pi /4})(z - 2e^{5i \pi/4})(z - 2e^{7 \pi i / 4})$ has simple zeroes at $2e^{i \pi / 4}, 2e^{3 i \pi / 4}, 2e^{5 i \pi / 4}, 2e^{7 i \pi / 4}$ so $f$ has simple zeroes at each of those points.
   3. $z^4 + 2z^2 + 1 = (z^2 + 1)^2 = (z+i)^2 (z-i)^2$ has zeroes of order two at $i$ and $-i$ so $f$ has poles of order two at $i$ and $-i$.
   4. $z^2 + z - 1 = \left( z - \dfrac{-1 - \sqrt{5}}{2} \right) \left( z - \dfrac{-1 + \sqrt{5}}{2} \right)$ has simple zeroes so $f$ has simple poles at $\dfrac{-1 - \sqrt{5}}{2}$ and $\dfrac{-1 + \sqrt{5}}{2}$.
6. 1. Since \[ \sin(1/z) = \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} \dfrac{1}{z^n} \] has infinitely many non-zero coefficients the singularity is essential.
   2. The function $g(z) = (\sin(z))^2$ is holomorphic on all of $\C$ so is represented by a power series \[ (\sin(z))^2 = \sum_{n=0}^\infty a_n z^n \] with infinite radius of convergence. However $(\sin z)^2$ has a zero of order two at the origin: the coefficients $a_0$ and $a_1$ are both zero whereas $a_2$ is non-zero. Indeed, calculating derivatives \[ ((\sin z)^2)' = 2 (\sin z)(\cos z) \qquad ((\sin z)^2)'' = 2 (\cos z)^2 - 2 (\sin z)^2 \] we see that the first derivative is zero at the origin while the second is non-zero at the origin. Thus \begin{align*} \dfrac{1}{z^3} (\sin z)^2 &= \dfrac{1}{z^3} \left( a_2 z^2 + a_3 z^3 + \cdots \right) \\ &= \dfrac{a_2}{z} + a_3 + a_4 z + a_5 z^2 + \cdots \end{align*} with $a_2$ non-zero so we have a simple pole at the origin.
   3. From our power series for cosine \begin{align*} \dfrac{1 - \cos(z)}{z^2} &= \dfrac{1}{z^2} \left( 1 - \left( 1 - \dfrac{z^2}{2!} + \dfrac{z^4}{4!} - \dfrac{z^6}{6!} + \cdots \right) \right) \\ & = \frac{1}{z^2} \left( \dfrac{z^2}{2!} - \dfrac{z^4}{4!} + \dfrac{z^6}{6!} - \cdots \right) \\ & = \dfrac{1}{2!} - \dfrac{z^2}{4!} + \dfrac{z^4}{6!} - \cdots \end{align*} so the singularity at the origin is removable.
7. Fix $M > 0$ such that $|f(z)| \le M$ for all $z \in D$. Fix $R > 0$ such that $\ball(b,R) \subset D$. Since $f$ is holomorphic on $\Ann(b,0,R)$ it has a Laurent series thereon. From Laurent's theorem \[ a_{-n} = \dfrac{1}{2 \pi i} \int\limits_\gamma \dfrac{f(z)}{(z-b)^{-n+1}} \intd z \] where $\gamma(t) = b + re^{it}$ for every $0 \l r \l R$ and every $n \in \N$. Thus \begin{align*} |a_{-n}| & = \dfrac{1}{2 \pi} \left| \int\limits_\gamma \dfrac{f(z)}{(z-b)^{-n+1}} \intd z \right| \\ & \le \dfrac{1}{2 \pi} \, 2 \pi r \, \dfrac{B}{r^{-n+1}} = B r^n \end{align*} from our estimation lemma. Since $r > 0$ was arbitrary we conclude that $a_{-n} = 0$ for all $n \in \N$. Thus the singularity is removable.