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Sagot :
To find the value of the contour integral
[tex]\[ \oint_{\partial B_{3 / 2}(2 i)} \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \cdot \frac{1}{(z-2 i)} \, dz, \][/tex]
we'll use principles from complex analysis, specifically leveraging the residue theorem.
1. Identify the Contour and Pole:
- The contour is [tex]\( \partial B_{3/2}(2i) \)[/tex], which is a circle centered at [tex]\( 2i \)[/tex] with radius [tex]\( 3/2 \)[/tex].
- The function inside the integral is [tex]\( \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \cdot \frac{1}{(z-2i)} \)[/tex].
2. Analyze the Function:
- The function can be split into two parts:
[tex]\[ \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \][/tex]
and
[tex]\[ \frac{1}{(z-2i)}. \][/tex]
- The term [tex]\( \frac{1}{(z-2i)} \)[/tex] clearly indicates a potential pole at [tex]\( z = 2i \)[/tex].
3. Finding the Residue:
- The Residue Theorem tells us that if we have a contour integral around a closed loop, the integral is [tex]\( 2\pi i \)[/tex] times the sum of residues inside the contour.
- First, we focus on finding the residue at [tex]\( z = 2i \)[/tex].
We need to calculate the residue at [tex]\( z = 2i \)[/tex] for [tex]\( \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \cdot \frac{1}{(z-2i)} \)[/tex].
4. Calculate Residue at [tex]\( z = 2i \)[/tex]:
- To find the residue, we need to isolate the singular part of the function.
- Consider [tex]\( f(z) = \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \)[/tex]. Evaluating this at [tex]\( z = 2i \)[/tex]:
[tex]\[ \operatorname{Re}\left(\frac{e^{\pi (2i)}-3}{2i}\right) \][/tex]
- Using Euler's formula, [tex]\( e^{\pi (2i)} = e^{2\pi i} = \cos(2\pi) + i\sin(2\pi) = 1 \)[/tex], giving us:
[tex]\[ \operatorname{Re}\left(\frac{1-3}{2i}\right) = \operatorname{Re}\left(\frac{-2}{2i}\right) = \operatorname{Re}\left(\frac{-2}{2i}\right) = \operatorname{Re}\left(\frac{-1}{i}\right) = \operatorname{Re}(i) = 0 \][/tex]
5. Evaluate the Integral:
- Since the real part calculated results in zero, there is no residue at [tex]\( z = 2i \)[/tex].
- According to the residue theorem, if the sum of residues within the contour is zero, the contour integral itself is zero.
Hence, the result of the integral is:
[tex]\[ \oint_{\partial B_{3/2}(2i)} \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \cdot \frac{1}{(z-2i)} \, dz = 0. \][/tex]
[tex]\[ \oint_{\partial B_{3 / 2}(2 i)} \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \cdot \frac{1}{(z-2 i)} \, dz, \][/tex]
we'll use principles from complex analysis, specifically leveraging the residue theorem.
1. Identify the Contour and Pole:
- The contour is [tex]\( \partial B_{3/2}(2i) \)[/tex], which is a circle centered at [tex]\( 2i \)[/tex] with radius [tex]\( 3/2 \)[/tex].
- The function inside the integral is [tex]\( \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \cdot \frac{1}{(z-2i)} \)[/tex].
2. Analyze the Function:
- The function can be split into two parts:
[tex]\[ \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \][/tex]
and
[tex]\[ \frac{1}{(z-2i)}. \][/tex]
- The term [tex]\( \frac{1}{(z-2i)} \)[/tex] clearly indicates a potential pole at [tex]\( z = 2i \)[/tex].
3. Finding the Residue:
- The Residue Theorem tells us that if we have a contour integral around a closed loop, the integral is [tex]\( 2\pi i \)[/tex] times the sum of residues inside the contour.
- First, we focus on finding the residue at [tex]\( z = 2i \)[/tex].
We need to calculate the residue at [tex]\( z = 2i \)[/tex] for [tex]\( \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \cdot \frac{1}{(z-2i)} \)[/tex].
4. Calculate Residue at [tex]\( z = 2i \)[/tex]:
- To find the residue, we need to isolate the singular part of the function.
- Consider [tex]\( f(z) = \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \)[/tex]. Evaluating this at [tex]\( z = 2i \)[/tex]:
[tex]\[ \operatorname{Re}\left(\frac{e^{\pi (2i)}-3}{2i}\right) \][/tex]
- Using Euler's formula, [tex]\( e^{\pi (2i)} = e^{2\pi i} = \cos(2\pi) + i\sin(2\pi) = 1 \)[/tex], giving us:
[tex]\[ \operatorname{Re}\left(\frac{1-3}{2i}\right) = \operatorname{Re}\left(\frac{-2}{2i}\right) = \operatorname{Re}\left(\frac{-2}{2i}\right) = \operatorname{Re}\left(\frac{-1}{i}\right) = \operatorname{Re}(i) = 0 \][/tex]
5. Evaluate the Integral:
- Since the real part calculated results in zero, there is no residue at [tex]\( z = 2i \)[/tex].
- According to the residue theorem, if the sum of residues within the contour is zero, the contour integral itself is zero.
Hence, the result of the integral is:
[tex]\[ \oint_{\partial B_{3/2}(2i)} \operatorname{Re}\left(\frac{e^{\pi z}-3}{z}\right) \cdot \frac{1}{(z-2i)} \, dz = 0. \][/tex]
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