Question

evaluate close integrals :
closed integral at c 1/(z^4+1) dz with contour c:|z|=4
closed integral c 1/(z^4+10z+9) dz with c:|z|=2
closed integral c e^z/(z^4+5z^3) dz with c:|z|=2

Answer 1

$\displaystyle (a)\\ \int_{|z| = 4} \frac{1}{z^4+1}\, \mathrm{d}z\\ \frac{1}{z^4+1}\,\, \textsf{possesses four simple poles at}\\ z = e^{\frac{j\pi}{4}}, e^{\frac{j3\pi}{4}}, e^{5\frac{\pi}{4}}, e^{j7\frac{\pi}{4}}\\ \textsf{of which all four are inside the circle}\, |z| = 4.\\ \begin{aligned} \textsf{The residue at}\, z = e^{\frac{j\pi}{4}} &\textsf{is}\, \lim_{z \rightarrow e^{\frac{j\pi}{4}}} \left( \left(z - e^{\frac{j\pi}{4}}\right) \times \frac{1}{1 + z^4}\right) \\&= \lim_{z \rightarrow e^{\frac{j\pi}{4}}} \left(\frac{1}{4z^3}\right)\,\, \textsf{by L'Hopital's rule} \\&= \frac{e^{\frac{-3j\pi}{4}}}{4} \end{aligned}\\ \begin{aligned} \textsf{The residue at}\, z = e^{\frac{j3\pi}{4}} &\textsf{is}\, \lim_{z \rightarrow e^{\frac{j3\pi}{4}}} \left( \left(z - e^{\frac{j3\pi}{4}}\right) \times \frac{1}{1 + z^4}\right) \\&= \lim_{z \rightarrow e^{\frac{\pi}{4}}} \left(\frac{1}{4z^3}\right)\,\, \textsf{by L'Hopital's rule} \\&= \frac{e^{\frac{-9j\pi}{4}}}{4} \end{aligned}\\ \begin{aligned} \textsf{The residue at}\, z = e^{\frac{j5\pi}{4}} &\textsf{is}\, \lim_{z \rightarrow e^{\frac{j5\pi}{4}}} \left( \left(z - e^{\frac{j5\pi}{4}}\right) \times \frac{1}{1 + z^4}\right) \\&= \lim_{z \rightarrow e^{\frac{j5\pi}{4}}} \left(\frac{1}{4z^3}\right)\,\, \textsf{by L'Hopital's rule} \\&= \frac{e^{\frac{-15j\pi}{4}}}{4} \end{aligned}\\ \begin{aligned} \textsf{The residue at}\, z = e^{\frac{j7\pi}{4}} &\textsf{is}\, \lim_{z \rightarrow e^{\frac{j7\pi}{4}}} \left( \left(z - e^{\frac{j7\pi}{4}}\right) \times \frac{1}{1 + z^4}\right) \\&= \lim_{z \rightarrow e^{\frac{j7\pi}{4}}} \left(\frac{1}{4z^3}\right)\,\, \textsf{by L'Hopital's rule} \\&= \frac{e^{\frac{-21j\pi}{4}}}{4} \end{aligned}\\ \begin{aligned} \int_{|z| = 4} \frac{1}{z^4+1}\, \mathrm{d}z &= 2\pi j \times \left(\frac{e^{\frac{-3j\pi}{4}}}{4} + \frac{e^{\frac{-9j\pi}{4}}}{4} + \frac{e^{\frac{-15j\pi}{4}}}{4} + \frac{e^{\frac{-21j\pi}{4}}}{4}\right) \\&= 2\pi e^{\frac{j\pi}{2}} \times \left(\frac{e^{\frac{-3j\pi}{4}}}{4} + \frac{e^{\frac{-9j\pi}{4}}}{4} + \frac{e^{\frac{-15j\pi}{4}}}{4} + \frac{e^{\frac{-21j\pi}{4}}}{4}\right) \\&= 2\pi \left(\frac{e^{\frac{-j\pi}{4}}}{4} + \frac{e^{\frac{-j7\pi}{4}}}{4} + \frac{e^{\frac{-13j\pi}{4}}}{4} + \frac{e^{\frac{-j19\pi}{4}}}{4}\right) \\&= \frac{\pi}{2}\left( e^{\frac{-j\pi}{4}} + e^{\frac{-j7\pi}{4}} + e^{\frac{-j13\pi}{4}} + e^{\frac{-j19\pi}{4}}\right) \\&= \frac{\pi}{2}\left( \cos\left(\frac{\pi}{4}\right) - j\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{7\pi}{4}\right) - j\sin\left(\frac{7\pi}{4}\right) + \right.\\ &\left. \cos\left(\frac{13\pi}{4}\right) - j\sin\left(\frac{13\pi}{4}\right) + \cos\left(\frac{19\pi}{4}\right) - j\sin\left(\frac{19\pi}{4}\right)\right) \\&= \frac{\pi}{2}\left(\cos\left(\frac{\pi}{4}\right) - j\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) - j\sin\left(\frac{\pi}{4}\right) \right.\\&\left.- \cos\left(\frac{\pi}{4}\right) - j\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) - j\sin\left(\frac{\pi}{4}\right)\right) \\&= \frac{\pi}{2}\left(- 4j\sin\left(\frac{\pi}{4}\right)\right) = -2\pi j \times \frac{\sqrt{2}}{2} = -j\pi\sqrt{2} \end{aligned}\\ (b)\\ \int_{|z| = 2} \frac{1}{z^4+10z+9}\, \mathrm{d}z = \frac{1}{z^4+10z+9}\,\, \textsf{possesses four simple poles at}\\ z = -1, -1.66, 1.33 - j1.91, 1.33 + j1.91\\ \textsf{of which all four are inside the circle}\, |z| = 2.\\ \begin{aligned} \textsf{The residue at}\, z = -1 \,&\textsf{is}\, \lim_{z \rightarrow -1} \left( \left(z + 1\right) \times \frac{1}{z^4+10z+9}\right) \\&= \lim_{z \rightarrow -1} \left(\frac{1}{z^3 - z^2 + z + 9}\right) \\&= \frac{1}{6} \end{aligned}\\ \begin{aligned} \textsf{The residue at}\, z = -1.66 \,&\textsf{is}\, \lim_{z \rightarrow -1.66} \left( \left(z + 1.66\right) \times \frac{1}{z^4+10z+9}\right) \\&= \lim_{z \rightarrow -1.66} \left(\frac{1}{4z^3 + 10}\right) \\&= \frac{1}{4(-1.66)^3 + 10} = -0.12 \end{aligned}\\ \begin{aligned} \textsf{The residue at}\, z = 1.33 - j1.91\,&\textsf{is}\, \lim_{z \rightarrow 1.33 - j1.91} \left( \left(z - 1.33 + j1.91\right) \times \frac{1}{z^4+10z+9}\right) \\&= \lim_{z \rightarrow 1.33 - j1.91} \left(\frac{1}{4z^3 + 10}\right) \\&= \frac{1}{4(1.33 - j1.91)^3 + 10} = -0.023 + j7.60 \end{aligned}\\ \begin{aligned} \textsf{The residue at}\, z = 1.33 + j1.91\,&\textsf{is}\, \lim_{z \rightarrow 1.33 + j1.91} \left( \left(z - 1.33 - j1.91\right) \times \frac{1}{z^4+10z+9}\right) \\&= \lim_{z \rightarrow 1.33 + j1.91} \left(\frac{1}{4z^3 + 10}\right) \\&= \frac{1}{4(1.33 + j1.91)^3 + 10} = -0.023 - j7.60 \end{aligned}\\ \begin{aligned} \therefore \int_{|z| = 2} \frac{1}{z^4+10z+9}\, \mathrm{d}z &= 2\pi j(\textsf{sum of residues}) \\&= 2\pi j \times 0.0006667 = j0.00133\pi\\ \end{aligned}\\ (c)\\ \int_{|z| = 2} \frac{e^z}{z^4+5z^3}\, \mathrm{d}z \\ \frac{e^z}{z^4+5z^3}\,\,\textsf{possesses two simple poles at}\\ z = 0, -5\\ \textsf{of which only the first is inside the circle}\, |z| = 2.\\ \begin{aligned} \textsf{The residue at}\, z = 0\,&\textsf{is}\, \lim_{z \rightarrow 0} \left(\frac{1}{(3 - 1)!}\frac{\mathrm{d}^2}{\mathrm{d}z^2} (z - 0)^3\times \frac{e^z}{z^4+5z^3}\right) \\&= \frac{1}{2}\lim_{z \rightarrow 0} \left(\frac{\mathrm{d}^2}{\mathrm{d}z^2}\frac{e^z}{z + 5}\right) \\&= \frac{1}{2}\lim_{z \rightarrow 0}\left( \frac{e^z}{z + 5} - \frac{2e^z}{(z + 5)^2} + \frac{2e^z}{(z + 5)^3}\right) \\&= \frac{1}{2}\left(\frac{1}{5} - \frac{2}{25} + \frac{2}{125}\right) =\frac{17}{125} \end{aligned}\\ \begin{aligned} \therefore \int_{|z| = 2} \frac{e^z}{z^4+5z^3}\, \mathrm{d}z &= 2\pi j \times \frac{17}{2(125)} \\&= \frac{17j\pi}{125} \end{aligned}$