VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end:Question 8 (1 point) cos(3 — 2] Weare toa thevatseot [| —— SED __g lo pA) A(2 + A(z + 4° over various contours C which are al understood to be traversed once inthe positive direction. In al cases, we want to use the (generalized) Cauchy Integral Formula, fo which the crucial step is hays the evaluation of certain function, or one ofits derivatives, at a certain point (besides, on stil needs to take the factor 2m into account). Match which procedure will be used for which contour. Careful, aptions can occur more an once, or mot at a 1. Evaluate the first derivative of cos(3 — 2) JE E— p() (2 + H(z +48 0 2. Evalue cos(3 — 2) Ce the unit circ [EE ac EE SP) (2 1 20) + 9) C = the circle with radius 3 atze2i - around -5 3. Evaluate the second dervative of [rm ee C= PEA) +4) —] © - the circle with racius 2 (And don't forget an extra factor of around 1-2 2! somewhere.) [Creve vertices 1, 11, and i 4. Bale | = the circle with radius 3 cos(3 — 2) around 0 P(E) (z— 20) (z + AF C = the square with atze-2i vertices 3,411.5 3nd 4 k 5 The integral over this contour cannot immedistely be solved with the CIF. (A deformation of the contour might help though.) 6. Evaluate the first dervative of cos(3— 2) . EL E— PEEP
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give answer in 2 step with explanation at the end of each step and final answer at the end:
Question 8 (1 point)
cos(3 — 2]
Weare toa thevatseot [| —— SED __g
lo pA) A(2 + A(z + 4°
over various contours C which are al understood to be traversed once inthe
positive direction. In al cases, we want to use the (generalized) Cauchy Integral
Formula, fo which the crucial step is hays the evaluation of certain function, or
one ofits derivatives, at a certain point (besides, on stil needs to take the factor
2m into account). Match which procedure will be used for which contour. Careful,
aptions can occur more an once, or mot at a
1. Evaluate the first derivative of
cos(3 — 2)
JE E—
p() (2 + H(z +48
0
2. Evalue
cos(3 — 2)
Ce the unit circ [EE ac EE
SP) (2 1 20) + 9)
C = the circle with radius 3 atze2i
- around -5
3. Evaluate the second dervative of
[rm ee C=
PEA) +4)
—] © - the circle with racius 2 (And don't forget an extra factor of
around 1-2 2! somewhere.)
[Creve
vertices 1, 11, and i 4. Bale
| = the circle with radius 3 cos(3 — 2)
around 0 P(E) (z— 20) (z + AF
C = the square with atze-2i
vertices 3,411.5 3nd 4
k 5 The integral over this contour
cannot immedistely be solved with
the CIF. (A deformation of the
contour might help though.)
6. Evaluate the first dervative of
cos(3— 2) .
EL E—
PEEP
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Asked by: Babu
Created at: 2025-07-12 12:02:02
Asked by: Babu
Created at: 2025-07-12 12:02:02
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Let's solve this in two steps, with explanations at each step:
---
### **Step 1: Identify the singularities for each contour**
The integrand has singularities at:
- \(z=0\) (simple pole, due to \(z^2\) in denominator)
- \(z = \pm 2i\) (simple poles, due to \((z^2 + 4)\))
- \(z = -4\) (third order pole, due to \((z+4)^3\))
We need to see which singularities are inside each contour:
1. **C = the unit circle**: Centered at 0, radius 1.
- Contains \(z=0\) only.
2. **C = the circle with radius 3 around -5**: Centered at -5, radius 3.
- Closest singularity is \(z=-4\). The circle goes from -8 to -2 on the real axis. So, it contains \(z=-4\).
3. **C = the triangle with vertices 1+i, 10i, and -5**:
- This is a large triangle. Contains \(z=0\), \(\pm 2i\), and possibly \(z=-4\).
4. **C = the circle with radius 2 around 1-2i**: Centered at \(1-2i\), radius 2.
- Closest singularity is \(z=0\) and \(z=2i\), but these are not inside (distance from center to 0 is about \(\sqrt{5}\)).
- Doesn't enclose any singularity.
5. **C = the triangle with vertices -1, 1-i, and 4i**:
- Let's check the coverage:
- All vertices have real parts between -1 and +1, so probably only \(z=0\) and maybe \(z=2i\) are close.
- But 4i is outside the range of the poles at \(z=2i\). So perhaps only \(z=0\).
6. **C = the circle with radius 3 around 0**: Centered at 0, radius 3.
- Contains \(z=0\), \(z=2i\), \(z=-2i\) (since \(|2i|=2<3\)).
- Does **not** contain \(z=-4\).
7. **C = the square with vertices -3, -4+i, -5, and -4-i**:
- All real parts between -3 and -5, so contains \(z=-4\) for sure.
---
### **Step 2: Match the procedures**
Now, match the given options based on which singularity is inside each contour:
#### 1. **C = the unit circle**
- Contains \(z=0\) (a double pole, due to \(z^2\) in denominator).
- The relevant formula is the **first derivative** at \(z=0\).
- **Match: Option 6** (Evaluate the first derivative... at \(z=0\) for \(z^2\) in denominator.)
#### 2. **C = the circle with radius 3 around -5**
- Contains \(z=-4\) (triple pole at \(z=-4\)).
- Use the **second derivative** at \(z=-4\), with an extra factor.
- **Match: Option 3** (Evaluate the second derivative... at \(z=-4\).)
#### 3. **C = the triangle with vertices 1+i, 10i, and -5**
- Contains \(z=0\), \(z=2i\), \(z=-2i\), and likely \(z=-4\).
- Too many poles; can't use CIF directly.
- **Match: Option 5** (The integral over this contour cannot immediately be solved...)
#### 4. **C = the circle with radius 2 around 1-2i**
- Does **not** contain any singularity.
- Integral is zero by Cauchy's theorem.
- **Match: Option 5** (The integral over this contour cannot immediately be solved...)
#### 5. **C = the triangle with vertices -1, 1-i, and 4i**
- Likely contains only \(z=0\). So, use the first derivative at \(z=0\) with \(z^2\) in the denominator.
- **Match: Option 6** (Evaluate the first derivative... at \(z=0\) for \(z^2\) in denominator.)
#### 6. **C = the circle with radius 3 around 0**
- Contains \(z=0\), \(z=2i\), \(z=-2i\).
- For \(z=2i\) and \(z=-2i\), these are simple poles.
- **Match: Option 2 and 4** (Evaluate at \(z=2i\) and at \(z=-2i\).)
#### 7. **C = the square with vertices -3, -4+i, -5, and -4-i**
- Contains \(z=-4\) (triple pole at \(z=-4\)).
- Use the **second derivative** at \(z=-4\), with an extra factor.
- **Match: Option 3** (Evaluate the second derivative... at \(z=-4\).)
---
## **Final Answers**
Here are the matches for each contour:
1. **C = the unit circle:**
**Option 6** (Evaluate the first derivative ... at \(z=0\) for \(z^2\) in denominator.)
2. **C = the circle with radius 3 around -5:**
**Option 3** (Evaluate the second derivative ... at \(z=-4\).)
3. **C = the triangle with vertices 1+i, 10i, and -5:**
**Option 5** (The integral over this contour cannot immediately be solved ...)
4. **C = the circle with radius 2 around 1-2i:**
**Option 5** (The integral over this contour cannot immediately be solved ...)
5. **C = the triangle with vertices -1, 1-i, and 4i:**
**Option 6** (Evaluate the first derivative ... at \(z=0\) for \(z^2\) in denominator.)
6. **C = the circle with radius 3 around 0:**
**Option 2 and 4** (Evaluate at \(z=2i\) and at \(z=-2i\), respectively.)
7. **C = the square with vertices -3, -4+i, -5, and -4-i:**
**Option 3** (Evaluate the second derivative ... at \(z=-4\).)
---
**Explanation:**
- The matching is based on which poles lie inside the given contour and the order of the pole at that location.
- For higher order poles, you use derivatives as specified by the Cauchy Integral Formula for derivatives.
- If multiple poles or no poles are inside, the standard CIF does not immediately apply.
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