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: Rewrite using Cauchy's Integral Formula.
The integral has the form $\oint_C \dfrac{(z+1)^2}{(z-i)(z-2)}\,dz$, with two singular points at $z=i$ and $z=2$. Only a singularity that sits inside the contour contributes to the integral.

Step 2: Test $z=2$ against the contour.
The contour is the circle $|z-(2+i)|=1.5$, centered at $2+i$. For $z=2$, the distance to the center is $|2-(2+i)|=|-i|=1$, and since $1<1.5$, this point sits inside the circle.

Step 3: Test $z=i$ against the contour.
For $z=i$, the distance to the center is $|i-(2+i)|=|-2|=2$, and since $2>1.5$, this point sits outside the circle. So the pole at $z=i$ plays no role here.

Step 4: Write $f(z)$ as $g(z)/(z-2)$ and use the formula directly.
Split off the singular factor: $g(z)=\dfrac{(z+1)^2}{z-i}$, so that $f(z)=\dfrac{g(z)}{z-2}$. Cauchy's Integral Formula gives $\oint_C \dfrac{g(z)}{z-2}\,dz=2\pi i\,g(2)$.

Step 5: Evaluate $g(2)$.
$g(2)=\dfrac{(2+1)^2}{2-i}=\dfrac{9}{2-i}$. Multiplying top and bottom by the conjugate $2+i$: $g(2)=\dfrac{9(2+i)}{(2-i)(2+i)}=\dfrac{9(2+i)}{5}=3.6+1.8i$.

Step 6: Multiply by $2\pi i$.
$2\pi i(3.6+1.8i)=2\pi(3.6i+1.8i^2)=2\pi(-1.8+3.6i)=-3.6\pi+7.2\pi i$.

Step 7: Take the modulus directly.
Here the real part is $-3.6\pi$ and the imaginary part is $7.2\pi$, so the modulus is $\pi\sqrt{3.6^2+7.2^2}=\pi\sqrt{12.96+51.84}=\pi\sqrt{64.8}$.

Step 8: Compute the number.
$\sqrt{64.8}\approx8.0499$, so the modulus is about $\pi\times8.0499\approx25.29$.
\[\boxed{25.29}\]