( a ) ∫ ∣ z ∣ = 4 1 z 4 + 1 d z 1 z 4 + 1 possesses four simple poles at z = e j π 4 , e j 3 π 4 , e 5 π 4 , e j 7 π 4 of which all four are inside the circle ∣ z ∣ = 4. The residue at z = e j π 4 is lim z → e j π 4 ( ( z − e j π 4 ) × 1 1 + z 4 ) = lim z → e j π 4 ( 1 4 z 3 ) by L’Hopital’s rule = e − 3 j π 4 4 The residue at z = e j 3 π 4 is lim z → e j 3 π 4 ( ( z − e j 3 π 4 ) × 1 1 + z 4 ) = lim z → e π 4 ( 1 4 z 3 ) by L’Hopital’s rule = e − 9 j π 4 4 The residue at z = e j 5 π 4 is lim z → e j 5 π 4 ( ( z − e j 5 π 4 ) × 1 1 + z 4 ) = lim z → e j 5 π 4 ( 1 4 z 3 ) by L’Hopital’s rule = e − 15 j π 4 4 The residue at z = e j 7 π 4 is lim z → e j 7 π 4 ( ( z − e j 7 π 4 ) × 1 1 + z 4 ) = lim z → e j 7 π 4 ( 1 4 z 3 ) by L’Hopital’s rule = e − 21 j π 4 4 ∫ ∣ z ∣ = 4 1 z 4 + 1 d z = 2 π j × ( e − 3 j π 4 4 + e − 9 j π 4 4 + e − 15 j π 4 4 + e − 21 j π 4 4 ) = 2 π e j π 2 × ( e − 3 j π 4 4 + e − 9 j π 4 4 + e − 15 j π 4 4 + e − 21 j π 4 4 ) = 2 π ( e − j π 4 4 + e − j 7 π 4 4 + e − 13 j π 4 4 + e − j 19 π 4 4 ) = π 2 ( e − j π 4 + e − j 7 π 4 + e − j 13 π 4 + e − j 19 π 4 ) = π 2 ( cos ( π 4 ) − j sin ( π 4 ) + cos ( 7 π 4 ) − j sin ( 7 π 4 ) + cos ( 13 π 4 ) − j sin ( 13 π 4 ) + cos ( 19 π 4 ) − j sin ( 19 π 4 ) ) = π 2 ( cos ( π 4 ) − j sin ( π 4 ) + cos ( π 4 ) − j sin ( π 4 ) − cos ( π 4 ) − j sin ( π 4 ) − cos ( π 4 ) − j sin ( π 4 ) ) = π 2 ( − 4 j sin ( π 4 ) ) = − 2 π j × 2 2 = − j π 2 ( b ) ∫ ∣ z ∣ = 2 1 z 4 + 10 z + 9 d z = 1 z 4 + 10 z + 9 possesses four simple poles at z = − 1 , − 1.66 , 1.33 − j 1.91 , 1.33 + j 1.91 of which all four are inside the circle ∣ z ∣ = 2. The residue at z = − 1 is lim z → − 1 ( ( z + 1 ) × 1 z 4 + 10 z + 9 ) = lim z → − 1 ( 1 z 3 − z 2 + z + 9 ) = 1 6 The residue at z = − 1.66 is lim z → − 1.66 ( ( z + 1.66 ) × 1 z 4 + 10 z + 9 ) = lim z → − 1.66 ( 1 4 z 3 + 10 ) = 1 4 ( − 1.66 ) 3 + 10 = − 0.12 The residue at z = 1.33 − j 1.91 is lim z → 1.33 − j 1.91 ( ( z − 1.33 + j 1.91 ) × 1 z 4 + 10 z + 9 ) = lim z → 1.33 − j 1.91 ( 1 4 z 3 + 10 ) = 1 4 ( 1.33 − j 1.91 ) 3 + 10 = − 0.023 + j 7.60 The residue at z = 1.33 + j 1.91 is lim z → 1.33 + j 1.91 ( ( z − 1.33 − j 1.91 ) × 1 z 4 + 10 z + 9 ) = lim z → 1.33 + j 1.91 ( 1 4 z 3 + 10 ) = 1 4 ( 1.33 + j 1.91 ) 3 + 10 = − 0.023 − j 7.60 ∴ ∫ ∣ z ∣ = 2 1 z 4 + 10 z + 9 d z = 2 π j ( sum of residues ) = 2 π j × 0.0006667 = j 0.00133 π ( c ) ∫ ∣ z ∣ = 2 e z z 4 + 5 z 3 d z e z z 4 + 5 z 3 possesses two simple poles at z = 0 , − 5 of which only the first is inside the circle ∣ z ∣ = 2. The residue at z = 0 is lim z → 0 ( 1 ( 3 − 1 ) ! d 2 d z 2 ( z − 0 ) 3 × e z z 4 + 5 z 3 ) = 1 2 lim z → 0 ( d 2 d z 2 e z z + 5 ) = 1 2 lim z → 0 ( e z z + 5 − 2 e z ( z + 5 ) 2 + 2 e z ( z + 5 ) 3 ) = 1 2 ( 1 5 − 2 25 + 2 125 ) = 17 125 ∴ ∫ ∣ z ∣ = 2 e z z 4 + 5 z 3 d z = 2 π j × 17 2 ( 125 ) = 17 j π 125 \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} ( a ) ∫ ∣ z ∣ = 4 z 4 + 1 1 d z z 4 + 1 1 possesses four simple poles at z = e 4 jπ , e 4 j 3 π , e 5 4 π , e j 7 4 π of which all four are inside the circle ∣ z ∣ = 4. The residue at z = e 4 jπ is z → e 4 jπ lim ( ( z − e 4 jπ ) × 1 + z 4 1 ) = z → e 4 jπ lim ( 4 z 3 1 ) by L’Hopital’s rule = 4 e 4 − 3 jπ The residue at z = e 4 j 3 π is z → e 4 j 3 π lim ( ( z − e 4 j 3 π ) × 1 + z 4 1 ) = z → e 4 π lim ( 4 z 3 1 ) by L’Hopital’s rule = 4 e 4 − 9 jπ The residue at z = e 4 j 5 π is z → e 4 j 5 π lim ( ( z − e 4 j 5 π ) × 1 + z 4 1 ) = z → e 4 j 5 π lim ( 4 z 3 1 ) by L’Hopital’s rule = 4 e 4 − 15 jπ The residue at z = e 4 j 7 π is z → e 4 j 7 π lim ( ( z − e 4 j 7 π ) × 1 + z 4 1 ) = z → e 4 j 7 π lim ( 4 z 3 1 ) by L’Hopital’s rule = 4 e 4 − 21 jπ ∫ ∣ z ∣ = 4 z 4 + 1 1 d z = 2 πj × ( 4 e 4 − 3 jπ + 4 e 4 − 9 jπ + 4 e 4 − 15 jπ + 4 e 4 − 21 jπ ) = 2 π e 2 jπ × ( 4 e 4 − 3 jπ + 4 e 4 − 9 jπ + 4 e 4 − 15 jπ + 4 e 4 − 21 jπ ) = 2 π ( 4 e 4 − jπ + 4 e 4 − j 7 π + 4 e 4 − 13 jπ + 4 e 4 − j 19 π ) = 2 π ( e 4 − jπ + e 4 − j 7 π + e 4 − j 13 π + e 4 − j 19 π ) = 2 π ( cos ( 4 π ) − j sin ( 4 π ) + cos ( 4 7 π ) − j sin ( 4 7 π ) + cos ( 4 13 π ) − j sin ( 4 13 π ) + cos ( 4 19 π ) − j sin ( 4 19 π ) ) = 2 π ( cos ( 4 π ) − j sin ( 4 π ) + cos ( 4 π ) − j sin ( 4 π ) − cos ( 4 π ) − j sin ( 4 π ) − cos ( 4 π ) − j sin ( 4 π ) ) = 2 π ( − 4 j sin ( 4 π ) ) = − 2 πj × 2 2 = − jπ 2 ( b ) ∫ ∣ z ∣ = 2 z 4 + 10 z + 9 1 d z = z 4 + 10 z + 9 1 possesses four simple poles at z = − 1 , − 1.66 , 1.33 − j 1.91 , 1.33 + j 1.91 of which all four are inside the circle ∣ z ∣ = 2. The residue at z = − 1 is z → − 1 lim ( ( z + 1 ) × z 4 + 10 z + 9 1 ) = z → − 1 lim ( z 3 − z 2 + z + 9 1 ) = 6 1 The residue at z = − 1.66 is z → − 1.66 lim ( ( z + 1.66 ) × z 4 + 10 z + 9 1 ) = z → − 1.66 lim ( 4 z 3 + 10 1 ) = 4 ( − 1.66 ) 3 + 10 1 = − 0.12 The residue at z = 1.33 − j 1.91 is z → 1.33 − j 1.91 lim ( ( z − 1.33 + j 1.91 ) × z 4 + 10 z + 9 1 ) = z → 1.33 − j 1.91 lim ( 4 z 3 + 10 1 ) = 4 ( 1.33 − j 1.91 ) 3 + 10 1 = − 0.023 + j 7.60 The residue at z = 1.33 + j 1.91 is z → 1.33 + j 1.91 lim ( ( z − 1.33 − j 1.91 ) × z 4 + 10 z + 9 1 ) = z → 1.33 + j 1.91 lim ( 4 z 3 + 10 1 ) = 4 ( 1.33 + j 1.91 ) 3 + 10 1 = − 0.023 − j 7.60 ∴ ∫ ∣ z ∣ = 2 z 4 + 10 z + 9 1 d z = 2 πj ( sum of residues ) = 2 πj × 0.0006667 = j 0.00133 π ( c ) ∫ ∣ z ∣ = 2 z 4 + 5 z 3 e z d z z 4 + 5 z 3 e z possesses two simple poles at z = 0 , − 5 of which only the first is inside the circle ∣ z ∣ = 2. The residue at z = 0 is z → 0 lim ( ( 3 − 1 )! 1 d z 2 d 2 ( z − 0 ) 3 × z 4 + 5 z 3 e z ) = 2 1 z → 0 lim ( d z 2 d 2 z + 5 e z ) = 2 1 z → 0 lim ( z + 5 e z − ( z + 5 ) 2 2 e z + ( z + 5 ) 3 2 e z ) = 2 1 ( 5 1 − 25 2 + 125 2 ) = 125 17 ∴ ∫ ∣ z ∣ = 2 z 4 + 5 z 3 e z d z = 2 πj × 2 ( 125 ) 17 = 125 17 jπ
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