Answer to Question #137714 in Complex Analysis for Usman

Question

Find residue
1)- 4/1+z^2
2)- 1/1-e^z
3)- sin2z/z^6

Answer 1

$\displaystyle(1) \\\textsf{Residue of} \hspace{0.1cm} f_1(z) = \frac{1}{z^2 + 1} \\ \displaystyle f(z)\hspace{0.1cm}\textsf{has two simple poles at}\hspace{0.1cm}z = \pm j ,\textsf{where}\hspace{0.1cm} j \hspace{0.1cm}\textsf{is a complex number}\\ \textsf{Calculating the residues at each poles.}\\ \textsf{The residue at} \hspace{0.1cm}z = j \hspace{0.1cm}\textsf{is}\\ \begin{aligned} \lim_{z \rightarrow j} \left(\frac{1}{(z – j)(z + j)}\cdot (z – j)\right) &= \lim_{z \rightarrow j} \left(\frac{1}{z+j}\right) \\ &= \lim_{z \rightarrow j} \frac{1}{2j} = \frac{1}{2j} = \frac{-j}{2} \end{aligned}\\ \displaystyle\textsf{The residue at} \hspace{0.1cm} z = -j \hspace{0.1cm}\textsf{is}\\ \begin{aligned} \lim_{z \rightarrow -j} \left(\frac{1}{(z – j)(z + j)}\cdot (z – j)\right) &= \lim_{z \rightarrow -j} \left(\frac{1}{z - j}\right) \\&= \lim_{z \rightarrow -j} \frac{1}{-2j} = \frac{1}{-2j} = \frac{j}{2} \end{aligned} \\ (2)\\ \displaystyle\textsf{Residue of} \hspace{0.1cm} f_2(z) = \frac{1}{1 - e^z} \\ f_2(z)\hspace{0.1cm}\textsf{has a pole when}\hspace{0.1cm} e^z = 1,\\ \implies z = 2jn\pi, \hspace{0.2cm} \forall n \in \mathbb{Z} \\ \therefore f_2(z)\hspace{0.1cm}\textsf{has poles at}\hspace{0.1cm} z = 2jn\pi\\ \textsf{Deriving the Laurent's series of}\hspace{0.1cm} f_2(z) \\ \displaystyle\begin{aligned} e^z - 1 &=\sum_{n=0}^\infty \frac{z^n}{n!} - 1 = \sum_{n=1}^\infty \frac{z^n}{n!} \\ &= z\sum_{n=1}^\infty \frac{z^{n - 1}}{n!} = z\sum_{n=0}^\infty \frac{z^n}{(n + 1)!} \\ &= z\left(1 + \sum_{n=1}^\infty \frac{z^n}{(n + 1)!}\right) \end{aligned} \\ \begin{aligned} \frac{1}{e^z - 1} &= \frac{1}{z}\left(1 + \sum_{n=1}^\infty \frac{z^n}{(n + 1)!}\right)^{-1} \\&= \frac{1}{z}\sum_{k=0}^\infty (-1)^k \left(\sum_{n=1}^\infty \frac{z^n}{(n + 1)!}\right)^k \\&= \frac{1}{z} - \frac{1}{z}\left(\frac{z}{2!} + \frac{z^2}{3!} +...\right) \\&+ \frac{1}{z}\left(\frac{z}{2!} + \frac{z^2}{3!} +...\right)^2 \\&- \frac{1}{z}\left(\frac{z}{2!} + \frac{z^2}{3!} +...\right)^3 \\&+ \frac{1}{z}\left(\frac{z}{2!} + \frac{z^2}{3!} +...\right)^4+... \\&= \frac{1}{z} - \frac{1}{2!} + \left(-\frac{1}{3!} + \frac{1}{(2!)^2}\right)z \\&+ \left(-\frac{1}{4!} + \frac{2}{2! \cdot 3!} - \frac{1}{(2!)^3}\right)z^2 \\&+ \left(-\frac{1}{5!} + \frac{2}{2! \cdot 4!} + \frac{1}{(3!)^2} - \frac{3}{(2!)^2\cdot 3!} + \frac{1}{(2!)^4}\right) z^3 +... \\&= \frac{1}{z} - \frac{1}{2} + \frac{z}{12} - \frac{z^3}{720} +... \end{aligned}\\ \displaystyle\textsf{By Cauchy's theorem,}\hspace{0.1cm} \frac{1}{2j\pi} \int_C \frac{1}{z^n} \, \mathrm{d}z = \begin{cases} =0, &\textsf{for}\hspace{0.1cm} n = 2, 3, 4,...\\ = 2\pi j, &\textsf{for}\hspace{0.1cm} n= 1\\ = 0,&\textsf{for}\hspace{0.1cm} n = -1,-2,-3,-4,... \end{cases} \\ \displaystyle\begin{aligned} \therefore Res\left(\frac{1}{1 - e^z}, z = 2jn\pi\right) \\&= -\frac{1}{2j\pi}\int_C \left(\frac{1}{z} - \frac{1}{2} + \frac{z}{12} - \frac{z^3}{720} +...\right) \mathrm{d}z\\&= -\frac{2 \pi j}{2 \pi j} = - 1 \end{aligned} \\ \displaystyle(3) \\\textsf{Residue of} \hspace{0.1cm} f_3(z) = \frac{\sin(2z)}{z^6} \\ f(z)\hspace{0.1cm}\textsf{has a pole at}\hspace{0.1cm} z = 0\\ \textsf{The Taylor series of}\hspace{0.1cm} \sin(2z) \hspace{0.1cm} \textsf{is}\\ \sin{(2z)} = (2z) - \frac{(2z)^3}{3!} + \frac{(2z)^5}{5!} -\frac{(2z)^7}{7!}+ O(z^9)\\ \frac{\sin(2z)}{z^6}= \frac{2}{z^5} - \frac{8}{6z^3} + \frac{4}{15z} - \frac{8z}{315} + O(z^3)\\ Res\left(\frac{1}{z^n}, z=0\right) = \frac{1}{2j\pi} \int_C \frac{1}{z^n} \, \mathrm{d}z \hspace{0.1cm} \textsf{By definition} \\ \textsf{By Cauchy's theorem,}\hspace{0.1cm} \frac{1}{2j\pi} \int_C \frac{1}{z^n} \, \mathrm{d}z = \begin{cases} =0, &\textsf{for}\hspace{0.1cm} n = 2, 3, 4,...\\ = 2\pi j, &\textsf{for}\hspace{0.1cm} n= 1\\ = 0,&\textsf{for}\hspace{0.1cm} n = -1,-2,-3,-4,... \end{cases}\\ \therefore Res(f_3(z), z=0)) = 0 - 0 + \frac{4}{15}\cdot\frac{2\pi j}{2 \pi j} - 0 + 0 - 0 = \frac{4}{15}$