Question

Let $C$ be the circle $|z| = \frac{3}{2}$ in the complex plane that is oriented in the counter-clockwise direction. The value of $a$ for which $\int_C \left(\frac{z+1}{z^2 - 3z + 2} + \frac{a}{z-1}\right) dz = 0$ is _______

• A. 1
• B. -1
• C. 2
• D. -2

Hint:

Poles that lie outside the contour of integration have no effect on the value of the line integral.
Focusing only on the singularities inside the boundary ($|z| < 1.5$) simplifies the calculation.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks us to find the value of the parameter $a$ such that the contour integral of the given complex function along the circle $|z| = 1.5$ is equal to zero.

Step 2: Key Formula or Approach:
According to the Cauchy Residue Theorem, the contour integral of a function $f(z)$ along a closed curve $C$ is:
\[ \oint_C f(z) dz = 2\pi j \sum \text{Res}(f(z), z_k) \]
where $z_k$ are the poles of the function that lie inside the contour $C$.

Step 3: Detailed Explanation:

• The given contour $C$ is the circle $|z| = \frac{3}{2} = 1.5$ centered at the origin.

• The integrand is:
\[ f(z) = \frac{z+1}{z^2 - 3z + 2} + \frac{a}{z-1} \]

• Factor the denominator of the first term:
\[ z^2 - 3z + 2 = (z-1)(z-2) \]

• This allows us to rewrite the integrand as:
\[ f(z) = \frac{z+1}{(z-1)(z-2)} + \frac{a}{z-1} \]

• The poles of the function are located at the roots of the denominators:
\[ z = 1 \quad \text{and} \quad z = 2 \]

• Next, we determine which poles lie inside the contour $C$ ($|z| < 1.5$):

• - The pole $z = 1$ is inside the contour since $|1| = 1 < 1.5$.

• - The pole $z = 2$ is outside the contour since $|2| = 2 > 1.5$.

• Therefore, only the pole at $z = 1$ contributes to the value of the contour integral.

• Now, we calculate the residue of $f(z)$ at $z = 1$:
\[ \text{Res}(f(z), z=1) = \lim_{z \to 1} (z-1) f(z) \]
\[ \text{Res}(f(z), z=1) = \lim_{z \to 1} \left[ \frac{z+1}{z-2} + a \right] \]

• Substituting $z = 1$ into the limit:
\[ \text{Res}(f(z), z=1) = \frac{1+1}{1-2} + a = \frac{2}{-1} + a = a - 2 \]

• Applying the Cauchy Residue Theorem:
\[ \oint_C f(z) dz = 2\pi j \cdot \text{Res}(f(z), z=1) \]
\[ \oint_C f(z) dz = 2\pi j (a - 2) \]

• Since we are given that this integral is equal to zero:
\[ 2\pi j (a - 2) = 0 \implies a - 2 = 0 \implies a = 2 \]

Step 4: Final Answer
The value of $a$ is 2, which corresponds to option (C).