Question

The value of the contour integral, $\displaystyle \oint_{C}\left(\frac{z+2}{z^{2}+2z+2}\right)dz$, where the contour $C$ is $\left\{ z:\; \left| z+1-\tfrac{3}{2}j \right| = 1 \right\}$, taken in the counter clockwise direction, is

• A. $-\pi(1+j)$
• B. $\pi(1+j)$
• C. $\pi(1-j)$
• D. $-\pi(1-j)$

Hint:

For $\oint \frac{g(z)}{h(z)}dz$ with simple poles, use $\operatorname{Res}=\dfrac{g(z_k)}{h'(z_k)}$. Always check which poles lie inside the given circle before applying $2\pi j$ times the sum of residues.

Answer 1

The Correct Option is B

Step 1: Locate the poles of the function
We are integrating \( f(z) = \dfrac{z+2}{z^{2}+2z+2} \). Factor the denominator: \[ z^{2}+2z+2 = (z+1)^{2}+1 = 0 \;\;\Rightarrow\;\; z = -1 \pm j. \] So the poles are at \(z=-1+j\) and \(z=-1-j\).

Step 2: Identify which poles lie inside the given contour
The contour is the circle \(|z+1-\tfrac{3}{2}j|=1\). This means the circle is centered at \(-1+\tfrac{3}{2}j\) with radius 1.
- Distance from center to \(z=-1+j\): \(|(-1+j)-(-1+1.5j)| = | -0.5j | = 0.5<1 \Rightarrow\) inside.
- Distance from center to \(z=-1-j\): \(|(-1-j)-(-1+1.5j)| = | -2.5j | = 2.5>1 \Rightarrow\) outside.
Thus only the pole at \(z=-1+j\) is enclosed by the contour.

Step 3: Compute the residue at the enclosed pole
We can write \(f(z) = \dfrac{g(z)}{h(z)}\) with \(g(z)=z+2\), \(h(z)=z^{2}+2z+2\). For a simple pole at \(z_1\), \[ \operatorname{Res}(f;z_1) = \frac{g(z_1)}{h'(z_1)}. \] Here, \(h'(z) = 2z+2\). Substituting \(z_1 = -1+j\):
\[ \operatorname{Res}(f; -1+j) = \frac{(-1+j)+2}{2(-1+j)+2} = \frac{1+j}{2j}. \] Multiply numerator and denominator by \(-j\): \[ \frac{1+j}{2j} = \frac{(1+j)(-j)}{2j(-j)} = \frac{(1+j)(-j)}{-2j^{2}}. \] Since \(j^{2}=-1\), denominator = \(-2)(-1) = 2\). Numerator = \((1+j)(-j) = -j - j^{2} = -j+1 = 1-j\). So residue = \(\tfrac{1-j}{2}\).

Step 4: Apply the residue theorem
For a positively oriented (CCW) contour, \[ \oint_{C} f(z)\,dz = 2\pi j \times \operatorname{Res}(f; -1+j). \] Substitute the residue: \[ = 2\pi j \cdot \frac{1-j}{2} = \pi j(1-j). \] Simplify: \(\pi j(1-j) = \pi (j - j^{2}) = \pi (j+1).\)

Final Answer:
\[ \boxed{\pi(1+j)} \]