Answer to Question #145040 in Complex Analysis for Khansa

Question #145040
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
1
Expert's answer
2020-11-22T18:15:49-0500

(a)z=41z4+1dz1z4+1  possesses four simple poles atz=ejπ4,ej3π4,e5π4,ej7π4of which all four are inside the circlez=4.The residue atz=ejπ4islimzejπ4((zejπ4)×11+z4)=limzejπ4(14z3)  by L’Hopital’s rule=e3jπ44The residue atz=ej3π4islimzej3π4((zej3π4)×11+z4)=limzeπ4(14z3)  by L’Hopital’s rule=e9jπ44The residue atz=ej5π4islimzej5π4((zej5π4)×11+z4)=limzej5π4(14z3)  by L’Hopital’s rule=e15jπ44The residue atz=ej7π4islimzej7π4((zej7π4)×11+z4)=limzej7π4(14z3)  by L’Hopital’s rule=e21jπ44z=41z4+1dz=2πj×(e3jπ44+e9jπ44+e15jπ44+e21jπ44)=2πejπ2×(e3jπ44+e9jπ44+e15jπ44+e21jπ44)=2π(ejπ44+ej7π44+e13jπ44+ej19π44)=π2(ejπ4+ej7π4+ej13π4+ej19π4)=π2(cos(π4)jsin(π4)+cos(7π4)jsin(7π4)+cos(13π4)jsin(13π4)+cos(19π4)jsin(19π4))=π2(cos(π4)jsin(π4)+cos(π4)jsin(π4)cos(π4)jsin(π4)cos(π4)jsin(π4))=π2(4jsin(π4))=2πj×22=jπ2(b)z=21z4+10z+9dz=1z4+10z+9  possesses four simple poles atz=1,1.66,1.33j1.91,1.33+j1.91of which all four are inside the circlez=2.The residue atz=1islimz1((z+1)×1z4+10z+9)=limz1(1z3z2+z+9)=16The residue atz=1.66islimz1.66((z+1.66)×1z4+10z+9)=limz1.66(14z3+10)=14(1.66)3+10=0.12The residue atz=1.33j1.91islimz1.33j1.91((z1.33+j1.91)×1z4+10z+9)=limz1.33j1.91(14z3+10)=14(1.33j1.91)3+10=0.023+j7.60The residue atz=1.33+j1.91islimz1.33+j1.91((z1.33j1.91)×1z4+10z+9)=limz1.33+j1.91(14z3+10)=14(1.33+j1.91)3+10=0.023j7.60z=21z4+10z+9dz=2πj(sum of residues)=2πj×0.0006667=j0.00133π(c)z=2ezz4+5z3dzezz4+5z3  possesses two simple poles atz=0,5of which only the first is inside the circlez=2.The residue atz=0islimz0(1(31)!d2dz2(z0)3×ezz4+5z3)=12limz0(d2dz2ezz+5)=12limz0(ezz+52ez(z+5)2+2ez(z+5)3)=12(15225+2125)=17125z=2ezz4+5z3dz=2πj×172(125)=17jπ125\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}


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