Statement of Cauchy’s integral formula:
Let \(f\)be analytic everywhere within and on a simple closed contour\(C\), taken in the positive sense. If \({z_0}\) is any point interior to\(C\), then
\(f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)
Proof of the formula:
Since, \(f\)is analytic everywhere, which implies\(f\)is continuous, therefore there corresponds to any positive number\(\varepsilon \), however small, a positive number \(\delta \)such that
\(\left| {f(z) - f({z_0})} \right| < \varepsilon \)whenever \(\left| {z - {z_0}} \right| < \delta \) Eq. (1)
Let us choose a positive number \(\rho \)which is less than \(\delta \)and is so small that, the small positively oriented circle\(\left| {z - {z_0}} \right| = \rho \), denoted by \({C_0}\)is interior to\(C\), as shown in the following figure:
Then, Eq. (1) becomes
\(\left| {f(z) - f({z_0})} \right| < \varepsilon \)whenever \(\left| {z - {z_0}} \right| = \rho \) Eq. (2)
Now,\(\frac{{f(z)}}{{z - {z_0}}}\) is analytic in the closed region in between \(C\)and\({C_0}\), then
\(\int_C {\frac{{f(z)}}{{z - {z_0}}}dz = \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz} } \) Eq. (3)
(If \(C\)and \({C_0}\) denotes positively oriented simple closed contour, where \({C_0}\) is interior to \(C\)and \(f\)is analytic in the closed region consisting of these contours and all points between them, then \(\int_C {f(z)dz = \int_{{C_0}} {f(z)dz} } \) )Now, \(\int_C {\frac{{f(z)}}{{z - {z_0}}}dz - \int_{{C_0}} {\frac{{f({z_0})}}{{z - {z_0}}}dz} } \)
\( = \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - \int_{{C_0}} {\frac{{f({z_0})}}{{z - {z_0}}}dz} } \) from Eq.(3)
\( = \int_{{C_0}} {\frac{{f(z) - f({z_0})}}{{z - {z_0}}}dz} \)
Now, \(\int_{{C_0}} {\frac{{f({z_0})}}{{z - {z_0}}}dz} = f({z_0})\int_{{C_0}} {\frac{1}{{z - {z_0}}}dz} = f({z_0})2\pi i\), since \(\int_{{C_0}} {\frac{1}{{z - {z_0}}}dz} = 2\pi i\)
Now,\(\int_{{C_0}} {\frac{{f(z) - f({z_0})}}{{z - {z_0}}}dz} \)
\( = \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - } \int_{{C_0}} {\frac{{f({z_0})}}{{z - {z_0}}}dz} \)
\( = \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - f({z_0})2\pi i} \)
From Eq. (2) it is clear that \({C_0}\) is of length \(2\pi \rho \)
\(\therefore \left| {\int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - f({z_0})2\pi i} } \right|\left\langle {\frac{\varepsilon }{\rho }} \right.2\pi \rho \) , (Since,\(\left| {\int_C {f(z)dz} } \right|\left\langle {ML} \right.\), where closed contour\(C\) is of length \(L\), \(f\)is analytic within and on \(C\) and \(M\)is such that \(\left| {f\left( z \right)} \right| < M\) )
\( = \left| {\int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - f({z_0})2\pi i} } \right|\left\langle {2\pi \varepsilon } \right.\)
Since, left hand side of the inequality is a non-negative constant that is less than any arbitrary small positive number, therefore it must be equal to zero.
\(\therefore \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - f({z_0})2\pi i} = 0\)
\( \Rightarrow \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz = f({z_0})2\pi i} \)
\( \Rightarrow f({z_0}) = \frac{1}{{2\pi i}}\int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)
\( \Rightarrow f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)
This proves the theorem.
An important Note:
Cauchy’s integral formula can be extended to
\({f^n}({z_0}) = \frac{{n!}}{{2\pi i}}\int_C {\frac{{f(z)}}{{{{\left( {z - {z_0}} \right)}^{n + 1}}}}dz\,} \), \(n = 0,1,2,...\);
\({f^n}({z_0})\)denotes the \({n^{th}}\)derivative of \(f\)at \(z = {z_0}\)
Some problems and their solutions are given below to illustrate the utility of Cauchy’s integral formula:
Question 1: Evaluate \(\int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} \)where \(C\)is the positively oriented circle\(\left| z \right| = 2\).
Solution: Consider\(f(z) = \frac{z}{{(9 - {z^2})}}\), which is analytic and within and on the circle\(\left| z \right| = 2\). Now, the point \({z_0} = - i\)lies within the circle\(\left| z \right| = 2\).
Therefore, from Cauchy’s integral formula \(f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)
We get that \(f( - i) = \frac{1}{{2\pi i}}\int_C {\frac{{\frac{z}{{9 - {z^2}}}}}{{z - ( - i)}}dz\,} \).
\( \Rightarrow f( - i) = \frac{1}{{2\pi i}}\int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} \)
\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} = f( - i)2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} = \frac{{ - i}}{{9 - ( - {i^2})}}2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} = \frac{{ - i}}{{10}}2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} = \frac{{ - {i^2}}}{{10}}2\pi \)
\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} = \frac{1}{{10}}2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} = \frac{\pi }{5}\)
Therefore, value of the required integral is\(\frac{\pi }{5}\).
Question 2: Evaluate \(\int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} \)where \(C\)is the positively oriented boundary of the square \(x = \pm 2\) and\(y = \pm 2\).
Solution: Consider\(f(z) = {e^{ - z}}\), which is analytic on and within the boundary, given by the square \(x = \pm 2\) and \(y = \pm 2\) .
Now, the point \({z_0} = \frac{{\pi i}}{2}\)lies within the boundary\(C\).
Therefore, from Cauchy’s integral formula \(f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)
We get that \(f\left( {\frac{{\pi i}}{2}} \right) = \frac{1}{{2\pi i}}\int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} \)
\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} = f\left( {\frac{{\pi i}}{2}} \right)2\pi i\)
\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} = {e^{ - \frac{{\pi i}}{2}}}2\pi i\)
\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} = \left( {\cos \frac{\pi }{2} - i\sin \frac{\pi }{2}} \right)2\pi i\)
\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} = \left( {0 - i} \right)2\pi i\)
\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} = - 2\pi {i^2}\)
\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} = - 2\pi ( - 1)\)
\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} = 2\pi \)
Therefore, value of the required integral is\(2\pi \).
Question 3: Evaluate \(\int_C {\frac{1}{{{z^2} + 4}}dz\,} \)where \(C\) is the positively oriented circle \(\left| {z - i} \right| = 2\).
Solution: Consider \(f(z) = \frac{1}{{z + 2i}}\)which is analytic within and on the circle \(\left| {z - i} \right| = 2\),
Now, the point \({z_0} = 2i\)lies within the circle\(\left| {z - i} \right| = 2\).
Therefore, from Cauchy’s integral formula \(f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \) we get that \(f(2i) = \frac{1}{{2\pi i}}\int_C {\frac{{\frac{1}{{\left( {z + 2i} \right)}}}}{{z - 2i}}dz\,} \)
\( \Rightarrow f(2i) = \frac{1}{{2\pi i}}\int_C {\frac{1}{{{z^2} + 4}}dz\,} \)
\( \Rightarrow \int_C {\frac{z}{{{z^2} + 4}}dz\,} = f(2i)2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{{z^2} + 4}}dz\,} = \frac{1}{{2i + 2i}}2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{{z^2} + 4}}dz\,} = \frac{1}{{4i}}2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{{z^2} + 4}}dz\,} = \frac{\pi }{2}\)
Therefore, value of the required integral is\(\frac{\pi }{2}\).
Question 4: If \(C\) is the positively oriented circle \(\left| z \right| = 1\), then evaluate\(\int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,} \).
Solution: Consider \(f(z) = {e^{2z}}\)which is entire i.e. analytic everywhere.
Now, the point \({z_0} = 0\)lies within the circle\(\left| z \right| = 1\).
Therefore, from the formula\({f^n}({z_0}) = \frac{{n!}}{{2\pi i}}\int_{{C_0}} {\frac{{f(z)}}{{{{\left( {z - {z_0}} \right)}^{n + 1}}}}dz\,} \)and taking \(n = 3\)we get
\(f'''(0) = \frac{{3!}}{{2\pi i}}\int_C {\frac{{{e^{2z}}}}{{{{\left( {z - 0} \right)}^{3 + 1}}}}dz\,} \)
\( \Rightarrow f'''(0) = \frac{6}{{2\pi i}}\int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,} \)
\( \Rightarrow \int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,} = \frac{{2\pi i}}{6}f'''(0)\)
\( \Rightarrow \int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,} = \frac{{\pi i}}{3}{\left( {8{e^{2z}}} \right)_{z = 0}}\)
\( \Rightarrow \int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,} = \frac{{8\pi i}}{3}\)
Therefore, value of the require integral is\(\frac{{8\pi i}}{3}\).
Question 5: If \(C\) is a positively oriented circle is\(\left| z \right| = 2\) then evaluate\(\int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,} \).
Solution: Consider\(f(z) = z\), which is analytic on and within \(\left| z \right| = 2\) .
Now, the point \({z_0} = 1\)lies within the circle\(\left| z \right| = 2\).
Therefore, using the formula\({f^n}({z_0}) = \frac{{n!}}{{2\pi i}}\int_{{C_0}} {\frac{{f(z)}}{{{{\left( {z - {z_0}} \right)}^{n + 1}}}}dz\,} \)and taking \(n = 1\)we get
\(f'(1) = \frac{{1!}}{{2\pi i}}\int_C {\frac{z}{{{{\left( {z - 1} \right)}^{1 + 1}}}}dz\,} \)
\( \Rightarrow f'(1) = \frac{1}{{2\pi i}}\int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,} \)
\( \Rightarrow \int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,} = f'(1)2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,} = 1.2\pi i\)
\( \Rightarrow \int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,} = 2\pi i\)
Therefore, value of the required integral is\(2\pi i\).
So, we hope readers have grasped the overall concept of Cauchy’s integral formula and its proof. In addition, the questions and solutions on evaluation of integrals of analytic functions through closed contours will be beneficial to readers and will help in solving these types of problems.
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