Question

The magnitude of the contour integral
\[ \oint_C \left(\frac{(z+1)^2}{(z-i)(z-2)}\right) dz \] over the contour \(C: |z-2-i| = 3/2\) is (round off to two decimal places).
Note: \(z\) is a complex variable and \(i=\sqrt{-1}\).

Hint:

Check which poles lie inside the given contour before applying the residue theorem.

Answer 1

Correct Answer: 25.29

Step 1: Locate the poles of the integrand.
The function inside the contour integral is
\[ f(z)=\frac{(z+1)^2}{(z-i)(z-2)} \]
This has two simple poles, one at \(z=i\) and one at \(z=2\).

Step 2: Identify the contour.
The contour \(C\) is the circle
\[ |z-2-i|=\frac{3}{2} \]
This is a circle centered at \(z_0=2+i\) with radius \(\frac{3}{2}\).

Step 3: Check which poles lie inside the contour.
The distance from the center \(2+i\) to the pole \(z=i\) is
\[ |(2+i)-i|=|2|=2 \]
Since \(2 > \frac{3}{2}\), the pole at \(z=i\) lies outside the contour.
The distance from the center \(2+i\) to the pole \(z=2\) is
\[ |(2+i)-2|=|i|=1 \]
Since \(1 < \frac{3}{2}\), the pole at \(z=2\) lies inside the contour.
So only the pole at \(z=2\) contributes to the integral.

Step 4: Find the residue at \(z=2\).
Since \(z=2\) is a simple pole, the residue is
\[ \text{Res}_{z=2}f(z)=\lim_{z\to2}(z-2)f(z)=\frac{(z+1)^2}{z-i}\Big|_{z=2}=\frac{9}{2-i} \]
Rationalize the denominator:
\[ \frac{9}{2-i}=\frac{9(2+i)}{(2-i)(2+i)}=\frac{9(2+i)}{4+1}=\frac{9(2+i)}{5}=\frac{18+9i}{5} \]

Step 5: Apply the residue theorem.
By the residue theorem,
\[ \oint_C f(z)\,dz=2\pi i\times\text{Res}_{z=2}f(z)=2\pi i\times\frac{18+9i}{5} \]
Multiply out:
\[ =\frac{2\pi}{5}(18i+9i^2)=\frac{2\pi}{5}(18i-9)=\frac{2\pi}{5}(-9+18i) \]
So the value of the integral is
\[ -\frac{18\pi}{5}+i\frac{36\pi}{5} \]

Step 6: Find the magnitude.
The magnitude of a complex number \(a+ib\) is \(\sqrt{a^2+b^2}\). Here
\[ a=-\frac{18\pi}{5},\qquad b=\frac{36\pi}{5} \]
\[ |a+ib|=\frac{2\pi}{5}\sqrt{9^2+18^2}=\frac{2\pi}{5}\sqrt{81+324}=\frac{2\pi}{5}\sqrt{405} \]
Since \(\sqrt{405}=\sqrt{81\times5}=9\sqrt5\),
\[ |a+ib|=\frac{2\pi}{5}\times9\sqrt5=\frac{18\pi\sqrt5}{5} \]

Step 7: Compute the numerical value.
\[ \frac{18\pi\sqrt5}{5}\approx\frac{18\times3.1416\times2.2361}{5}\approx25.29 \]

Final Answer:
\[ \boxed{25.29} \]