Week 10 Worksheet - Solutions

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Poles and Residues

1. 1. $z(1-z^2) = z(1-z)(1+z)$ has simple zeroes at each of $-1,0,1$ so its reciprocal has simple poles at all three of those points. \begin{align*} \Res(f,-1) & = \lim_{z \to -1} (z - (-1)) \dfrac{1}{z(1-z^2)} = \lim_{z \to -1} \dfrac{1}{z(1-z)} = -\dfrac{1}{2} \\ \Res(f,0) & = \lim_{z \to 0} (z - 0) \dfrac{1}{z(1-z^2)} = \lim_{z \to 0} \dfrac{1}{1-z^2} = 1 \\ \Res(f,1) & = \lim_{z \to 1} (z - 1) \dfrac{1}{z(1-z^2)} = \lim_{z \to 1} \dfrac{-1}{z(1+z)} = -\dfrac{1}{2} \end{align*}
   2. Since $\sin$ is non-zero at every $\pi n + \frac{\pi}{2}$ and $\cos' = \sin$ we can say that the ratio has a simple pole at every $\pi n + \frac{\pi}{2}$. Now \[ \Res(f,n\pi + \tfrac{\pi}{2}) = \dfrac{\sin(n \pi + \tfrac{\pi}{2})}{- \sin(n \pi + \tfrac{\pi}{2})} = -1 \] using our lemma on residues of ratios at simple poles.
   3. The numerator is non-zero at each of the simple zeroes of the denominator so the function has a simple pole at each of these zeroes. Now \[ \Res(f,e^{ik \pi/4}) = \lim_{z \to e^{i k \pi /4}} (z - e^{i k \pi / 4}) \dfrac{z}{1 + z^4} = \dfrac{e^{i \pi / 4} }{ \displaystyle\prod_{j \ne k} (z + e^{ij\pi/4})} \] but we can calculate more easily that \[ \Res(f,e^{ik \pi/4}) = \dfrac{e^{ik \pi / 4}}{4 e^{3ik \pi / 4}} = \dfrac{1}{4e^{i k \pi / 2}} \] using our lemma about residues of ratios at simple poles.
   4. The denominator is $(z+i)^2(z-i)^2$ which has zeroes of order two at $i$ and $-i$. Since the numerator is non-zero at $i$ and $-i$ the ratio has poles of order two at both $i$ and $-i$. We calculate the residues using our lemma for residues at higher-order poles. \begin{align*} \Res(f,i) & = \lim_{z \to i} \left( \dfrac{(z-i)^2(z+1)^2}{(z^2 + 1)^2} \right)' \\ & = \lim_{z \to i} \left( \dfrac{(z+1)^2}{(z+i)^2} \right)' \\ & = \lim_{z \to i} \dfrac{2(z+1)(z+i)^2 - 2(z+i)(z+1)^2}{(z+i)^4} \\ & = \lim_{z \to i} \dfrac{2(z+1)(z+i) - 2(z+1)^2}{(z+i)^3} \\ & = \dfrac{2(1+i)2i - 2(i+1)^2}{(2i)^3} = - \dfrac{i}{2} \\ \Res(f,-i) & = \lim_{z \to -i} \left( \dfrac{(z-(-i))^2(z+1)^2}{(z^2 + 1)^2} \right)' \\ & = \lim_{z \to -i} \left( \dfrac{(z+1)^2}{(z-i)^2} \right)' \\ & = \lim_{z \to -i} \dfrac{2(z+1)(z-i)^2 - 2(z-i)(z+1)^2}{(z-i)^4} \\ & = \lim_{z \to -i} \dfrac{2(z+1)(z-i) - 2(z+1)^2}{(z-i)^3} \\ & = \dfrac{-2(1-i)2i - 2(i-1)^2}{(-2i)^3} = \dfrac{i}{2} \end{align*}
2. 1. From the Taylor series \[ \dfrac{\sin z}{z^2} = \dfrac{1}{z^2} \left( z - \dfrac{z^3}{3!} + \dfrac{z^5}{5!} - \cdots \right) = \dfrac{1}{z} - \dfrac{z}{3!} + \dfrac{z^3}{5} - \cdots \] has a simple pole at the origin with a residue of 1.
   2. First note that \[ (\sin z)^2 = \dfrac{1 - \cos(2z)}{2} = z^2 - \dfrac{z^4}{3} + \cdots \] so we can write \[ \dfrac{(\sin z)^2}{z^4} = \dfrac{1}{z^4} \left( z^2 - \dfrac{z^4}{3} + \cdots \right) = \dfrac{1}{z^2} - \dfrac{1}{3} + \cdots \] showing a pole of order two at the origin with a residue of $0$.
3. 1. On $\Ann(0,0,1)$ we have \[ \dfrac{1}{(1-z)^2} = \left( \dfrac{1}{1-z} \right)' = \left( \sum_{n=0}^\infty z^n \right)' = \sum_{n=0}^\infty (n+1) z^n \] so $f(z) = \dfrac{1}{z} + 2 + 3z + 4z^2 + \cdots$ thereon. Similarly, on $\Ann(1,0,1)$ we have \[ \dfrac{1}{z} = \dfrac{1}{1 - (1-z)} = \sum_{n=0}^\infty (-1)^n (z-1)^n \] so $f(z) = \dfrac{1}{(z-1)^2} - \dfrac{1}{(z-1)} + 1 - (z-1) + (z-1)^2 - \cdots$ thereon.
   2. From the above series $f$ has a simple pole at $0$ and a pole of order two at $1$. The residue at the origin is 1 and the residue at 1 is $-1$.
   3. Since the numerator is never zero and $z (1-z)^2$ has a simple zero at $0$; a zero of order two at $1$; we conclude that $f$ has a simple pole at $0$ and a pole of order two at $1$. The residues are \[ \lim_{z \to 0} \dfrac{z}{z(1-z)^2} = 1 \] and \[ \lim_{z \to 1} \left( \dfrac{(z-1)^2}{z(1-z)^2} \right)' = \lim_{z \to 0} \left( \dfrac{1}{z} \right)' = \lim_{z \to 1} \dfrac{-1}{z^2} = -1 \] respectively, confirming our earlier answers.
4. We can write \begin{align*} f(z) = \dfrac{z^3 + 3z^2 + 3z + 1}{(z-1)^3} & = \dfrac{((z-1) + 1)^3 + 3((z-1) + 1)^2 + 3((z-1) + 1) + 1}{(z-1)^3} \\ & = \dfrac{(z-1)^3 + 6(z-1)^2 + 12(z-1) + 8}{(z-1)^3} \\ & = \dfrac{8}{(z-1)^3} + \dfrac{12}{(z-1)^2} + \dfrac{6}{(z-1)^1} + 1 \end{align*} so the residue at $1$ is 6.
5. The denominator has Taylor series \[ z^2 \sin(z) = z^2 \left( z - \dfrac{z^3}{3!} + \dfrac{z^5}{5!} - \cdots \right) = z^3 - \dfrac{z^5}{3!} + \dfrac{z^7}{5!} - \cdots \] so the order of the zero at the origin is three. Since the numerator is non-zero at the origin we have a pole of order three at the origin. We therefore (taking $m=2$ in our residue lemma) must calculate \begin{align*} & \lim_{z \to 0} \dfrac{ \left( (z-0)^3 \dfrac{1}{z^2 \sin(z)} \right)^{(2)}}{2} \\ = & \frac{1}{2} \lim_{z \to 0} \left( \dfrac{z}{\sin(z)} \right)^{(2)} \\ = & \dfrac{1}{2} \lim_{z \to 0} \left( \dfrac{\sin z - z \cos z}{(\sin z)^2} \right)' \\ = & \dfrac{1}{2} \lim_{z \to 0} \dfrac{(\cos z - (\cos z - z \sin z))(\sin z)^2 - 2 (\sin z)(\cos z) (\sin z - z \cos z)}{(\sin z)^4} \\ = & \dfrac{1}{2} \lim_{z \to 0} \dfrac{z (\sin z)^3 - 2(\sin z)^2 \cos z + 2 z (\sin z) (\cos z)^2}{(\sin z)^4} \end{align*} which has the form $0/0$ if we simply plug in $z = 0$. To proceed we must calculate the orders of the zeroes of the numerator and the denominator. \begin{align*} 2 (\sin z)^2 (\cos z) & = 2 z^2 \left( 1 - \dfrac{z^2}{3!} + \cdots \right)^2 \left( 1 - \dfrac{z^2}{2!} + \cdots \right) = 2z^2 - \dfrac{5}{3} z^4 + \cdots \\ 2 z (\sin z)(\cos z)^2 & = 2z^2 \left( 1 - \dfrac{z^2}{3!} + \cdots \right) \left( 1 - \dfrac{z^2}{2!} + \cdots \right)^2 = 2z^2 - \dfrac{7}{3} z^4 + \cdots \\ (\sin z)^4 & = \left( z - \dfrac{z^2}{3!} + \cdots \right)^4 = z^4 - \cdots \\ z (\sin z)^3 & = z \left( z - \dfrac{z^2}{3!} + \cdots \right)^4 = z^4 - \cdots \end{align*} so we can write \begin{align*} & \dfrac{1}{2} \lim_{z \to 0} \dfrac{z (\sin z)^3 - 2(\sin z)^2 \cos z + 2 z (\sin z) (\cos z)^2}{(\sin z)^4} \\ = & \dfrac{1}{2} \lim_{z \to 0} \dfrac{z^4 - \left( 2z^2 - \frac{5}{3} z^4 \right) + \left( 2z^2 - \frac{7}{3} z^4 \right)}{z^4} \\ = & \dfrac{1}{2} \lim_{z \to 0} \dfrac{\frac{1}{3} z^4}{z^4} = \dfrac{1}{6} \end{align*} finally.
6. We can write $f(z) = (z-b)^m F(z)$ and $g(z) = (z-b)^n G(z)$ where $F,G$ are holomorphic on some $\ball(b,r)$ and both $F(b), G(b)$ are non-zero. Thus \[ f(z) g(z) = (z-b)^{m+n} F(z) G(z) \] with $F(z) G(z)$ holomorphic on $\ball(b,r)$ and non-zero at $b$. The coefficient $a_0$ in the Taylor series of $F(z) G(z)$ at $b$ is therefore non-zero and from \[ f(z) g(z) = (z-b)^{m+n} \left( a_0 + a_1 (z-b) + a_2 (z-b)^2 + \cdots \right) \] we see that $f(z)g(z)$ has a zero of order $m+n$ at $b$.

Cauchy's Residue Theorem

7. 1. First factor $z^2 - 5z + 6 = (z-2)(z-3)$ to see that the integrand has simple poles at $2$ and $3$. Both are inside $\gamma_4$ and $\wind(\gamma_4,2) = 1 = \wind(\gamma_4,3)$ so \[ \int\limits_{\gamma_4} \dfrac{1}{z^2 - 5z + 6} \intd z = 2 \pi i \Res(f,2) + 2 \pi i \Res(f,3) \] by Cauchy's residue theorem. It remains to calculate the residues \begin{align*} \Res(f,2) & = \lim_{z \to 2} \dfrac{z-2}{(z-2)(z-3)} = -1 \\ \Res(f,3) & = \lim_{z \to 3} \dfrac{z-3}{(z-2)(z-3)} = 1 \\ \end{align*} giving $\displaystyle\int\limits_{\gamma_4} \dfrac{1}{z^2 - 5z + 6} \intd z = 0$.
   2. This is the same function as a) but the contour now only contains one of the poles: the simple pole at 2. Thus \[ \int\limits_{\gamma_4} \dfrac{1}{z^2 - 5z + 6} \intd z = 2 \pi i \Res(f,2) = - 2 \pi i \]
   3. Fix $a \in \R$. Since $\exp(az)$ is holomorphic and never zero, the integrand has simple poles at $i$ and $-i$. Both are inside $\gamma_2$. The residues are \begin{align*} \Res(f,i) & = \lim_{z \to i} \dfrac{\exp(az)(z-i)}{(z-i)(z+i)} = \dfrac{\exp(ai)}{2i} = \dfrac{\cos(a) + i \sin(a)}{2i} \\ \Res(f,-i) & = \lim_{z \to -i} \dfrac{\exp(az)(z+i)}{(z-i)(z+i)} = \dfrac{\exp(-ai)}{2i} = \dfrac{\cos(a) - i \sin(a)}{-2i} \end{align*} so Cauchy's residue theorem gives \[ \int\limits_{\gamma_2} \dfrac{\exp(az)}{1-z^2} = 2 \pi i \Res(f,i) + 2 \pi i \Res(f,-i) = 2 \pi i \sin(a) \] because both winding numbers are $1$.