( 1 ) Residue of f 1 ( z ) = 1 z 2 + 1 f ( z ) has two simple poles at z = ± j , where j is a complex number Calculating the residues at each poles. The residue at z = j is lim z → j ( 1 ( z – j ) ( z + j ) ⋅ ( z – j ) ) = lim z → j ( 1 z + j ) = lim z → j 1 2 j = 1 2 j = − j 2 The residue at z = − j is lim z → − j ( 1 ( z – j ) ( z + j ) ⋅ ( z – j ) ) = lim z → − j ( 1 z − j ) = lim z → − j 1 − 2 j = 1 − 2 j = j 2 ( 2 ) Residue of f 2 ( z ) = 1 1 − e z f 2 ( z ) has a pole when e z = 1 , ⟹ z = 2 j n π , ∀ n ∈ Z ∴ f 2 ( z ) has poles at z = 2 j n π Deriving the Laurent’s series of f 2 ( z ) e z − 1 = ∑ n = 0 ∞ z n n ! − 1 = ∑ n = 1 ∞ z n n ! = z ∑ n = 1 ∞ z n − 1 n ! = z ∑ n = 0 ∞ z n ( n + 1 ) ! = z ( 1 + ∑ n = 1 ∞ z n ( n + 1 ) ! ) 1 e z − 1 = 1 z ( 1 + ∑ n = 1 ∞ z n ( n + 1 ) ! ) − 1 = 1 z ∑ k = 0 ∞ ( − 1 ) k ( ∑ n = 1 ∞ z n ( n + 1 ) ! ) k = 1 z − 1 z ( z 2 ! + z 2 3 ! + . . . ) + 1 z ( z 2 ! + z 2 3 ! + . . . ) 2 − 1 z ( z 2 ! + z 2 3 ! + . . . ) 3 + 1 z ( z 2 ! + z 2 3 ! + . . . ) 4 + . . . = 1 z − 1 2 ! + ( − 1 3 ! + 1 ( 2 ! ) 2 ) z + ( − 1 4 ! + 2 2 ! ⋅ 3 ! − 1 ( 2 ! ) 3 ) z 2 + ( − 1 5 ! + 2 2 ! ⋅ 4 ! + 1 ( 3 ! ) 2 − 3 ( 2 ! ) 2 ⋅ 3 ! + 1 ( 2 ! ) 4 ) z 3 + . . . = 1 z − 1 2 + z 12 − z 3 720 + . . . By Cauchy’s theorem, 1 2 j π ∫ C 1 z n d z = { = 0 , for n = 2 , 3 , 4 , . . . = 2 π j , for n = 1 = 0 , for n = − 1 , − 2 , − 3 , − 4 , . . . ∴ R e s ( 1 1 − e z , z = 2 j n π ) = − 1 2 j π ∫ C ( 1 z − 1 2 + z 12 − z 3 720 + . . . ) d z = − 2 π j 2 π j = − 1 ( 3 ) Residue of f 3 ( z ) = sin ( 2 z ) z 6 f ( z ) has a pole at z = 0 The Taylor series of sin ( 2 z ) is sin ( 2 z ) = ( 2 z ) − ( 2 z ) 3 3 ! + ( 2 z ) 5 5 ! − ( 2 z ) 7 7 ! + O ( z 9 ) sin ( 2 z ) z 6 = 2 z 5 − 8 6 z 3 + 4 15 z − 8 z 315 + O ( z 3 ) R e s ( 1 z n , z = 0 ) = 1 2 j π ∫ C 1 z n d z By definition By Cauchy’s theorem, 1 2 j π ∫ C 1 z n d z = { = 0 , for n = 2 , 3 , 4 , . . . = 2 π j , for n = 1 = 0 , for n = − 1 , − 2 , − 3 , − 4 , . . . ∴ R e s ( f 3 ( z ) , z = 0 ) ) = 0 − 0 + 4 15 ⋅ 2 π j 2 π j − 0 + 0 − 0 = 4 15 \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} ( 1 ) Residue of f 1 ( z ) = z 2 + 1 1 f ( z ) has two simple poles at z = ± j , where j is a complex number Calculating the residues at each poles. The residue at z = j is z → j lim ( ( z – j ) ( z + j ) 1 ⋅ ( z – j ) ) = z → j lim ( z + j 1 ) = z → j lim 2 j 1 = 2 j 1 = 2 − j The residue at z = − j is z → − j lim ( ( z – j ) ( z + j ) 1 ⋅ ( z – j ) ) = z → − j lim ( z − j 1 ) = z → − j lim − 2 j 1 = − 2 j 1 = 2 j ( 2 ) Residue of f 2 ( z ) = 1 − e z 1 f 2 ( z ) has a pole when e z = 1 , ⟹ z = 2 jnπ , ∀ n ∈ Z ∴ f 2 ( z ) has poles at z = 2 jnπ Deriving the Laurent’s series of f 2 ( z ) e z − 1 = n = 0 ∑ ∞ n ! z n − 1 = n = 1 ∑ ∞ n ! z n = z n = 1 ∑ ∞ n ! z n − 1 = z n = 0 ∑ ∞ ( n + 1 )! z n = z ( 1 + n = 1 ∑ ∞ ( n + 1 )! z n ) e z − 1 1 = z 1 ( 1 + n = 1 ∑ ∞ ( n + 1 )! z n ) − 1 = z 1 k = 0 ∑ ∞ ( − 1 ) k ( n = 1 ∑ ∞ ( n + 1 )! z n ) k = z 1 − z 1 ( 2 ! z + 3 ! z 2 + ... ) + z 1 ( 2 ! z + 3 ! z 2 + ... ) 2 − z 1 ( 2 ! z + 3 ! z 2 + ... ) 3 + z 1 ( 2 ! z + 3 ! z 2 + ... ) 4 + ... = z 1 − 2 ! 1 + ( − 3 ! 1 + ( 2 ! ) 2 1 ) z + ( − 4 ! 1 + 2 ! ⋅ 3 ! 2 − ( 2 ! ) 3 1 ) z 2 + ( − 5 ! 1 + 2 ! ⋅ 4 ! 2 + ( 3 ! ) 2 1 − ( 2 ! ) 2 ⋅ 3 ! 3 + ( 2 ! ) 4 1 ) z 3 + ... = z 1 − 2 1 + 12 z − 720 z 3 + ... By Cauchy’s theorem, 2 jπ 1 ∫ C z n 1 d z = ⎩ ⎨ ⎧ = 0 , = 2 πj , = 0 , for n = 2 , 3 , 4 , ... for n = 1 for n = − 1 , − 2 , − 3 , − 4 , ... ∴ R es ( 1 − e z 1 , z = 2 jnπ ) = − 2 jπ 1 ∫ C ( z 1 − 2 1 + 12 z − 720 z 3 + ... ) d z = − 2 πj 2 πj = − 1 ( 3 ) Residue of f 3 ( z ) = z 6 sin ( 2 z ) f ( z ) has a pole at z = 0 The Taylor series of sin ( 2 z ) is sin ( 2 z ) = ( 2 z ) − 3 ! ( 2 z ) 3 + 5 ! ( 2 z ) 5 − 7 ! ( 2 z ) 7 + O ( z 9 ) z 6 sin ( 2 z ) = z 5 2 − 6 z 3 8 + 15 z 4 − 315 8 z + O ( z 3 ) R es ( z n 1 , z = 0 ) = 2 jπ 1 ∫ C z n 1 d z By definition By Cauchy’s theorem, 2 jπ 1 ∫ C z n 1 d z = ⎩ ⎨ ⎧ = 0 , = 2 πj , = 0 , for n = 2 , 3 , 4 , ... for n = 1 for n = − 1 , − 2 , − 3 , − 4 , ... ∴ R es ( f 3 ( z ) , z = 0 )) = 0 − 0 + 15 4 ⋅ 2 πj 2 πj − 0 + 0 − 0 = 15 4
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