Chapter 19

$$\int_{0}^{2 \pi} \frac{d \theta}{1+\frac{1}{2} \sin \theta}=\oint_{C} \frac{4}{z^{2}+4 i z-1} d z=(4) 2 \pi i \operatorname{Res}(f(z),(\sqrt{3}-2) i)=\frac{4 \pi}{\sqrt{3}}$$

$$\text { Using } e^{2 z}=\sum_{k=0}^{\infty} \frac{2^{k} z^{k}}{k !} \text { we obtain }$$ \(\frac{e^{2 z}-1}{z}=\frac{\left(1+\frac{2}{1 !} z+\frac{2^{2}}{2 !} z^{2}+\frac{2^{3}}{3 !} z^{3}+\cdots\right)-1}{z}=\frac{1}{z}\left(\frac{2}{1 !} z+\frac{2^{2}}{2 !} z^{2}+\frac{2^{3}}{3 !} z^{3}+\cdots\right)=\frac{2}{1 !}+\frac{2^{2}}{2 !} z+\frac{2^{3}}{3 !} z^{2}+\cdots\) From the form of the last series we see that \(z=0\) is a removable singularity. Define \(f(0)=2\)

$$\frac{z}{1+z}=z\left[1-z+z^{2}-z^{3}+\cdots\right]=z-z^{2}+z^{3}-z^{4}+\cdots=\sum_{k=1}^{\infty}(-1)^{k+1} z^{k} ; \quad R=1$$

$$f(z)=\frac{1}{z}\left(1-\frac{z^{2}}{2 !}+\frac{z^{4}}{4 !}-\frac{z^{6}}{6 !}+\cdots\right)=\frac{1}{z}-\frac{z}{2 !}+\frac{z^{3}}{4 !}-\frac{z^{5}}{6 !}+\cdots$$

Write \(f(z)=z^{4}-16=\left(z^{2}-4\right)\left(z^{2}+4\right)=(z-2)(z+2)(z-2 i)(z+2 i)\) to see that \(2,-2,2 i,\) and \(-2 i\) are zeros of \(f .\) Now \(f^{\prime}(z)=4 z^{3}\) and \(f^{\prime}(2) \neq 0, f^{\prime}(-2) \neq 0, f^{\prime}(2 i) \neq 0,\) and \(f^{\prime}(-2 i) \neq 0 .\) This indicates that each zero is of order one.

Using the binomial series gives $$\frac{z}{(1-z)}=z\left[1+3 z+\frac{3 \cdot 4}{2 !} z+\frac{3 \cdot 4 \cdot 5}{3 !} z^{3}+\cdots\right]=z+3 z^{2}+\frac{3 \cdot 4}{2 !} z^{3}+\frac{3 \cdot 4 \cdot 5}{3 !} z^{4}+\cdots \text { where } R=1$$

$$f(z)=e^{-2 / z^{2}}=\sum_{k=0}^{\infty} \frac{\left(-2 / z^{2}\right)^{k}}{k !}=\cdots-\frac{2^{3}}{3 ! z^{6}}+\frac{2^{2}}{2 ! z^{4}}-\frac{2}{1 ! z^{2}}+1 ; \quad \operatorname{Res}(f(z), 0)=0$$

Replacing \(z\) in \(e^{z}=\sum_{k=0}^{\infty} \frac{z^{k}}{k !}\) by \(-2 z\) gives \(e^{-2 z}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !}(2 z)^{k}\) where \(R=\infty\)

Write \(f(z)=z^{2}\left(z^{2}+1\right)=z^{2}(z-i)(z+i)\) to see that \(0, i,\) and \(-i\) are zeros of \(f .\) Now \(f^{\prime}(z)=4 z^{3}+2 z\) and \(f^{\prime}(i) \neq 0\) and \(f^{\prime}(-i) \neq 0 .\) This indicates that \(z=i\) and \(z=-i\) are zeros of order one. However \(f^{\prime}(0)=0,\) but \(f^{\prime \prime}(0)=2 \neq 0 .\) Hence \(z=0\) is a zero of order two.

$$f(z)=\frac{c \cdot c^{z-1}}{z-1}=\frac{c}{z-1}\left(1+\frac{(z-1)}{1 !}+\frac{(z-1)^{2}}{2 !}+\frac{(z-1)^{3}}{3 !}+\cdots\right)=\frac{c}{z-1}+\frac{c}{1 !}+\frac{c(z-1)}{2 !}+\frac{c(z-1)^{2}}{3 !}+\cdots$$

$$\begin{aligned} \int_{0}^{\pi} \frac{d \theta}{1+\sin ^{2} \theta} &=\frac{1}{2} \int_{0}^{2 \pi} \frac{d \theta}{1+\sin ^{2} \theta}=-\frac{2}{i} \oint_{C} \frac{z}{z^{4}-6 z^{2}+1} d z \\ &=\left(-\frac{2}{i}\right) 2 \pi i[\operatorname{Res}(f(z), \sqrt{3-2 \sqrt{2}})+\operatorname{Res}(f(z),-\sqrt{3-2 \sqrt{2}})]=\frac{\pi}{\sqrt{2}} \end{aligned}$$

Replacing \(z\) in \(e^{z}=\sum_{k=0}^{\infty} \frac{z^{k}}{k !} \quad\) by \(-z^{2}\) and multiplying the result by \(z\) gives \(z e^{-z^{2}}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !} z^{2 k+1}\) where \(R=\infty\)

Write \(f(z)=\left(z^{2}+9\right) / z=(z-3 i)(z+3 i) / z\) to see that \(3 i\) and \(-3 i\) are zeros of \(f .\) Now \(f^{\prime}(z)=1-9 / z^{2}\) and \(f^{\prime}(3 i)=f^{\prime}(-3 i)=2 \neq 0 .\) This indicates that each zero is of order one.

Converges. To see this write the general term as \(\left(\frac{2}{5}\right)^{n} \frac{1+n 2^{-n} i}{1+3 n 5^{-n} i}\).

Write \(f(z)=e^{z}\left(e^{z}-1\right)\) to see that \(2 n \pi i, n=0,\pm 1,\pm 2, \ldots\) are zeros of \(f .\) Now \(f^{\prime}(z)=2 e^{2 z}-e^{z}\) and \(f^{\prime}(2 n \pi i)=2 e^{4 n \pi i}-e^{2 n \pi i}=1 \neq 0 .\) This indicates that each zero is of order one.

Subtracting the series for \(e^{z}\) and \(e^{-z}\) gives \(\sinh z=\frac{1}{2}\left(e^{z}-e^{-z}\right)=\sum_{k=0}^{\infty} \frac{z^{2 k+1}}{(2 k+1) !}\) where \(R=\infty\)

Adding the series for \(e^{z}\) and \(e^{-z}\) gives \(\cosh z=\frac{1}{2}\left(e^{z}+e^{-z}\right)=\sum_{k=0}^{\infty} \frac{z^{2 k}}{(2 k) !}\) where \(R=\infty\)

The zeros of \(f\) are the zeros of \(\sin z,\) that is, \(n \pi, n=0,\pm 1,\pm 2, \ldots .\) From \(f^{\prime}(z)=2 \sin z \cos z\) we see \(f^{\prime}(n \pi)=0\) From \(f^{\prime \prime}(z)=2\left(-\sin ^{2} z+\cos ^{2} z\right)\) we see \(f^{\prime \prime}(n \pi) \neq 0 .\) This indicates that each zero is of order two.

$$f(z)=\frac{1}{z^{2}} \cdot \frac{1}{1-\frac{3}{z}}=\frac{1}{z^{2}}\left[1+\frac{3}{z}+\frac{3^{2}}{z^{2}}+\frac{3^{3}}{z^{3}}+\dots\right]=\frac{1}{z^{2}}+\frac{3}{z^{3}}+\frac{3^{2}}{z^{1}}+\frac{3^{3}}{z^{5}}+\cdots$$

From \(\quad f(z)=z\left(1-\cos z^{2}\right)=z\left(-\frac{z^{4}}{2 !}+\frac{z^{8}}{4 !}-\cdots\right)=z^{5}\left(-\frac{1}{2 !}+\frac{z^{4}}{4 !}-\cdots\right)\) we see that \(z=0\) is a zero of order five.

Replacing \(z\) in \(\cos z=\sum_{k=0}^{\infty}(-1)^{k} \frac{z^{2 k}}{(2 k) !}\) by \(z / 2\) gives \(\cos \frac{z}{2}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !}\left(\frac{z}{2}\right)^{2 k}\) where \(R=\infty\)

From \(\quad f(z)=z-\sin z=\frac{z^{3}}{3 !}-\frac{z^{5}}{5 !}+\cdots=z^{3}\left(\frac{1}{3 !}-\frac{z^{2}}{5 !}+\cdots\right)\) we see that \(z=0\) is a zero of order three.

Replacing \(z\) in \(\sin z=\sum_{k=0}^{\infty}(-1)^{k} \frac{z^{2 k+1}}{(2 k+1) !}\) by \(3 z\) gives \(\sin 3 z=\sum_{k=0}^{\infty}(-1)^{k} \frac{(3 z)^{2 k+1}}{(2 k+1) !}\) where \(R=\infty\)

$$\begin{array}{l} \operatorname{Res}(f(z),-1)=\lim _{z \rightarrow-1}(z+1) \cdot \frac{5 z^{2}-4 z+3}{(z+1)(z+2)(z+3)}=\lim _{z \rightarrow-1} \frac{5 z^{2}-4 z+3}{(z+2)(z+3)}=6 \\ \operatorname{Res}(f(z),-2)=\lim _{z \rightarrow-2}(z+2) \cdot \frac{5 z^{2}-4 z+3}{(z+1)(z+2)(z+3)}=\lim _{z \rightarrow-2} \frac{5 z^{2}-4 z+3}{(z+1)(z+3)}=-31 \\ \operatorname{Res}(f(z),-3)=\lim _{z \rightarrow-3}(z+3) \cdot \frac{5 z^{2}-4 z+3}{(z+1)(z+2)(z+3)}=\lim _{z \rightarrow-3} \frac{5 z^{2}-4 z+3}{(z+1)(z+2)}=30 \end{array}$$

$$\operatorname{Re}\left(z_{n}\right)=\frac{8 n^{2}+n}{4 n^{2}+1} \rightarrow 2 \text { as } n \rightarrow \infty, \text { and } \operatorname{Im}\left(z_{n}\right)=\frac{6 n^{2}-4 n}{4 n^{2}+1} \rightarrow \frac{3}{2} \text { as } n \rightarrow \infty$$.

From the series \(e^{z}=-\sum_{k=0}^{\infty} \frac{(z-\pi i)^{k}}{k !}\) centered at \(\pi i\) and \\[ \begin{aligned} f(z) &=1-\pi i+z+e^{z}=1-\pi i+z+\left(-1-\frac{z-\pi i}{1 !}-\frac{(z-\pi i)^{2}}{2 !}-\frac{(z-\pi i)^{3}}{3 !}-\cdots\right) \\ &=-\frac{(z-\pi i)^{2}}{2 !}-\frac{(z-\pi i)^{3}}{3 !}-\cdots=(z-\pi i)^{2}\left(-\frac{1}{2 !}-\frac{z-\pi i}{3 !}-\cdots\right) \end{aligned} \\] we see that \(z=\pi i\) is a zero of order two

$$\begin{aligned} f(z) &=\frac{1}{3}\left[\frac{1}{z-3}-\frac{1}{z}\right]=\frac{1}{3}\left[\frac{1}{-4+z+1}-\frac{1}{z+1-1}\right]=\frac{1}{3}\left[-\frac{1}{4} \cdot \frac{1}{1-\frac{z+1}{4}}-\frac{1}{z+1} \cdot \frac{1}{1-\frac{1}{z+1}}\right] \\ &=\frac{1}{3}\left[-\frac{1}{4}\left(1+\frac{z+1}{4}+\frac{(z+1)^{2}}{4^{2}}+\frac{(z+1)^{3}}{4^{3}}+\cdots\right)-\frac{1}{z+1}\left(1+\frac{1}{z+1}+\frac{1}{(z+1)^{2}}+\frac{1}{(z+1)^{3}}+\cdots\right)\right] \\\ &=\cdots-\frac{1}{(z+1)^{2}}-\frac{1}{z+1}-\frac{1}{12}-\frac{z+1}{3 \cdot 4^{2}}-\frac{(z+1)^{2}}{3 \cdot 4^{3}}-\cdots \end{aligned}$$

$$\begin{aligned}&\text { Write } z_{n}=\left(\frac{1}{4}+\frac{1}{4} i\right)^{n} \text { in polar form as } z_{n}=\left(\frac{\sqrt{2}}{4}\right)^{n} \cos n \theta+i\left(\frac{\sqrt{2}}{4}\right)^{n} \sin n \theta . \text { Now }\\\ &\operatorname{Re}\left(z_{n}\right)=\left(\frac{\sqrt{2}}{4}\right)^{n} \cos n \theta \rightarrow 0 \text { as } n \rightarrow \infty \quad \text { and } \quad \operatorname{Im}\left(z_{n}\right)=\left(\frac{\sqrt{2}}{4}\right)^{n} \sin n \theta \rightarrow 0 \text { as } n \rightarrow \infty\\\&\operatorname{since} \sqrt{2} / 4<1\end{aligned}$$

$$\begin{array}{l} \operatorname{Res}(f(z), 0)=\frac{1}{1 !} \lim _{z \rightarrow 0} \frac{d}{d z}\left[z^{2} \cdot \frac{\cos z}{z^{2}(z-\pi)^{3}}\right]=\lim _{z \rightarrow 0} \frac{-(z-\pi) \sin z-3 \cos z}{(z-\pi)^{4}}=-\frac{3}{\pi^{4}} \\\ \operatorname{Res}(f(z), \pi)=\frac{1}{2 !} \lim _{z \rightarrow \pi} \frac{d^{2}}{d z^{2}}\left[(z-\pi)^{3} \cdot \frac{\cos z}{z^{2}(z-\pi)^{3}}\right]=\frac{1}{2} \lim _{z \rightarrow \pi} \frac{-z^{2} \cos z+4 z \sin z+6 \cos z}{z^{4}}=\frac{\pi^{2}-6}{2 \pi^{4}} \end{array}$$

$$\begin{aligned}&S_{n}=\frac{1}{1+2 i}-\frac{1}{2+2 i}+\frac{1}{2+2 i}-\frac{1}{3+2 i}+\frac{1}{3+2 i}-\frac{1}{4+2 i}+\cdots+\frac{1}{n+2 i}-\frac{1}{n+1+2 i}=\frac{1}{1+2 i}-\frac{1}{n+1+2 i}\\\&\text { Thus, } \lim _{n \rightarrow \infty} S_{n}=\frac{1}{1+2 i}=\frac{1}{5}-\frac{2}{5} i \end{aligned}$$

$$\int_{-\infty}^{\infty} \frac{x}{\left(x^{2}+4\right)^{3}} d x=0 \quad(\text { The integrand is an odd function })$$

We identify \(a=4 i\) and \(z=1 / 3 .\) since \(|z|=1 / 3<1\) the series converges to \\[\frac{4 i}{1-1 / 3}=6 i\\]

We identify \(a=i / 2\) and \(z=i / 2 .\) since \(|z|=1 / 2<1\) the series converges to \\[\frac{i / 2}{1-i / 2}=-\frac{1}{5}+\frac{2}{5} i\\]

$$\begin{aligned} f(z) &=\frac{1}{3(z+1)}+\frac{2}{3} \cdot \frac{1}{(z+1)-3}=\frac{1}{3(z+1)}+\frac{2}{3(z+1)} \cdot \frac{1}{1-\frac{3}{z+1}} \\ &=\frac{1}{3(z+1)}+\frac{2}{3(z+1)}\left(1+\frac{3}{z+1}+\frac{3^{2}}{(z+1)^{2}}+\frac{3^{3}}{(z+1)^{3}}+\cdots\right) \\\ &=\frac{1}{z+1}+\frac{2}{(z+1)^{2}}+\frac{2 \cdot 3}{(z+1)^{3}}+\frac{2 \cdot 3^{2}}{(z+1)^{4}}+\cdots \end{aligned}$$

$$\int_{0}^{\infty} \frac{x^{2}+1}{x^{4}+1} d x=\frac{1}{2} \int_{-\infty}^{\infty} \frac{x^{2}+1}{x^{4}+1} d x=\pi i \operatorname{Res}(f(z), i)=\frac{\pi}{\sqrt{2}}$$

$$\begin{aligned} f(z) &=\frac{1 / 3}{z+1}+\frac{2 / 3}{z-2}=\frac{1}{3 z} \cdot \frac{1}{1+\frac{1}{z}}-\frac{1}{3} \cdot \frac{1}{1-\frac{z}{2}}=\frac{1}{3 z}\left(1-\frac{1}{z}+\frac{1}{z^{2}}-\frac{1}{z^{3}}+\cdots\right)-\frac{1}{3}\left(1+\frac{z}{2}+\frac{z^{2}}{2^{2}}+\frac{z^{3}}{2^{3}}+\cdots\right) \\\ &=\cdots-\frac{1}{3 z^{2}}+\frac{1}{3 z}-\frac{1}{3}-\frac{z}{3 \cdot 2}-\frac{z^{2}}{3 \cdot 2^{2}}-\cdots \end{aligned}$$

We identify \(a=3\) and \(z=2 /(1+2 i) .\) since \(|z|=2 / \sqrt{5}<1\) the series converges to \\[\frac{3}{1-\frac{2}{1+2 i}}=\frac{9}{5}-\frac{12}{5} i\\]

$$\int_{0}^{\infty} \frac{1}{x^{6}+1} d x=\frac{1}{2} \int_{-\infty}^{\infty} \frac{1}{x^{6}+1} d x=\pi i\left[\operatorname{Res}\left(f(z), \frac{\sqrt{3}}{2}+\frac{1}{2} i\right)+\operatorname{Res}(f(z), i)+\operatorname{Res}\left(f(z), \frac{-\sqrt{3}}{2}+\frac{1}{2} i\right)\right]=\frac{\pi}{3}$$

From the Laurent series \\[ f(z)=\frac{e^{z}}{z^{2}}=\frac{\left(1+\frac{z}{1 !}+\frac{z^{2}}{2 !}+\cdots\right)}{z^{2}}=\frac{1}{z^{2}}+\frac{1}{z}+\frac{1}{2 !}+\cdots \\] we see that 0 is a pole of order two.

\(z=0\) is a removable singularity of the function \((\sin z) / z .\) From \(f(z)=\frac{\sin z}{z(z-1)}\) we see that only 1 is a (simple) pole.

$$\oint_{C} \frac{z}{z^{4}-1} d z=2 \pi i[\operatorname{Res}(f(z),-1)+\operatorname{Res}(f(z), 1)+\operatorname{Res}(f(z),-i)+\operatorname{Res}(f(z), i)]=2 \pi i\left[\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}\right]=0$$

The function \(f(z)=\frac{\sin (1 / z)}{\cos (1 / z)}\) fails to be defined at \(z=0\) and at the solutions of \(\cos \frac{1}{z}=0,\) that is, at \(\frac{1}{z}=(2 n+1) \frac{\pi}{2}, n=0,\pm 1,\pm 2, \ldots .\) since \(z=\frac{2}{(2 n+1) \pi}, n=0,\pm 1,\pm 2, \ldots\) we see that in any neighborhood of \(z=0\) there are points at which \(f\) is not defined and thus not analytic. Hence \(z=0\) is a non-isolated singularity.

\(\int_{-\infty}^{\infty} \frac{e^{i x}}{\left(x^{2}+4\right)^{2}} d x=2 \pi i \operatorname{Res}(f(z), 2 i)=\frac{3 e^{-2}}{16} \pi ; \quad \int_{-\infty}^{\infty} \frac{\cos x}{\left(x^{2}+4\right)^{2}} d x=\operatorname{Re}\left(\int_{-\infty}^{\infty} \frac{e^{i x}}{\left(x^{2}+4\right)^{2}} d x\right)=\frac{3 e^{-2}}{16} \pi\) Therefore $$\int_{0}^{\infty} \frac{\cos x}{\left(x^{2}+4\right)^{2}} d x=\frac{1}{2}\left(\frac{3 e^{-2}}{16} \pi\right)=\frac{3 e^{-2}}{32} \pi$$

From the Laurent series \\[ f(z)=z^{3}\left[\frac{1}{z}-\frac{1}{3 !}\left(\frac{1}{z}\right)^{3}+\frac{1}{5 !}\left(\frac{1}{z}\right)^{5}-\frac{1}{7 !}\left(\frac{1^{7}}{z}\right)+\cdots\right]=z^{2}-\frac{1}{3 !}+\frac{1}{5 ! z^{2}}-\frac{1}{7 ! z^{4}}+\cdots, \quad 0<|z| \\] we see that the principal part contains an infinite number of nonzero terms. Hence \(z=0\) is an essential singularity.

From \\[\lim _{n \rightarrow \infty}\left|\frac{\frac{1}{(n+1)^{2}(3+4 i)^{n+1}}}{\frac{1}{n^{2}(3+4 i)^{n}}}\right|=\lim _{n \rightarrow \infty}\left(\frac{n}{n+1}\right)^{2} \frac{1}{|3+4 i|}=\frac{1}{5}\\] we see that the radius of convergence is \(R=5 .\) The circle of convergence is \(|z+3 i|=5\).

$$\int_{-\infty}^{\infty} \frac{e^{i x}}{x^{2}+4 x+5} d x=2 \pi i \operatorname{Res}(f(z),-2+i)=\pi e^{-1-2 i} . \text { Therefore }$$ $$\int_{-\infty}^{\infty} \frac{\sin x}{x^{2}+4 x+5} d x=\operatorname{Im}\left(\int_{-\infty}^{\infty} \frac{e^{i x}}{x^{2}+4 x+5} d x\right)=-\pi e^{-1} \sin 2$$

From \(\lim _{n \rightarrow \infty} \sqrt[n]{\left|\frac{1}{n^{n}}\right|}=\lim _{n \rightarrow \infty} \frac{1}{n}=0\) we see that the radius of convergence is \(\infty .\) The power series with center 0 converges absolutely for all \(z\).

$$\begin{aligned} \int_{-\infty}^{\infty} \frac{e^{2 i x}}{x^{4}+1} d x &=2 \pi i\left[\operatorname{Res}\left(f(z), \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i\right)+\operatorname{Res}\left(f(z),-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i\right)\right] \\ &=2 \pi i\left[\left(-\frac{\sqrt{2}}{8}-\frac{\sqrt{2}}{8} i\right) e^{(-\sqrt{2}+\sqrt{2} i)}+\left(\frac{\sqrt{2}}{8}-\frac{\sqrt{2}}{8} i\right) e^{(-\sqrt{2}-\sqrt{2} i)}\right] \\ &=\pi e^{-\sqrt{2}}\left[\frac{\sqrt{2}}{2} \cos \sqrt{2}+\frac{\sqrt{2}}{2} \sin \sqrt{2}\right] \end{aligned}$$ $$\int_{-\infty}^{\infty} \frac{\cos 2 x}{x^{4}+1} d x=\operatorname{Re}\left(\int_{-\infty}^{\infty} \frac{e^{2 i x}}{x^{4}+1} d x\right)=\pi e^{-\sqrt{2}}\left[\frac{\sqrt{2}}{2} \cos \sqrt{2}+\frac{\sqrt{2}}{2} \sin \sqrt{2}\right]$$ Therefore $$\int_{0}^{\infty} \frac{\cos 2 x}{x^{4}+1} d x=\pi e^{-\sqrt{2}} \frac{\sqrt{2}}{4}(\cos \sqrt{2}+\sin \sqrt{2})$$

The distance from \(2+5 i\) to \(i\) is \(|2+5 i-i|=|2+4 i|=2 \sqrt{5}\)

$$\oint_{C} \frac{\cot \pi z}{z^{2}} d z=2 \pi i \operatorname{Res}(f(z), 0)=2 \pi i\left(-\frac{\pi}{3}\right)=-\frac{2 \pi^{2}}{3} i$$ Note: \(z=0\) is a pole of order three. Use L'Hôpital's rule (or Mathematica ) to show that $$\operatorname{Res}(f(z), 0)=\frac{1}{2} \lim _{z \rightarrow 0} \frac{d^{2}}{d z^{2}} z \cot \pi z=\frac{1}{2} \lim _{z \rightarrow 0}\left[-2 \pi \csc ^{2} \pi z+2 \pi^{2} z \cot \pi z \csc ^{2} \pi z\right]=\frac{1}{2}\left(-\frac{2 \pi}{3}\right)=-\frac{\pi}{3}$$