Question

Let $r = \text{Min}\{\alpha, \beta, \gamma\}$, $R = \text{Max}\{\alpha, \beta, \gamma\}$, $f(z) = \frac{z}{(z-\alpha)(z-\beta)(z-\gamma)}$. $I_1 = \oint_{C_1} f(z)dz$ and $I_2 = \oint_{C_2} f(z)dz$, where $C_1 : |z| < r$ and $C_2 : |z| = R+1$, then $I_1 + I_2 = $}

• A. \(2\pi i\)
• B. \(0\)
• C. \(\pi i\)
• D. \(-\pi i\)

Hint:

For any rational function where the degree of the denominator exceeds the degree of the numerator by 2 or more ($\text{deg}(Q) \ge \text{deg}(P) + 2$), the integral over any closed contour enclosing all the poles is identically zero.

Answer 1

The Correct Option is B

Concept: This problem is evaluated using principles of complex integration, specifically Cauchy's Integral Theorem and the Residue Theorem. - Cauchy's Integral Theorem states that if a function $f(z)$ is analytic everywhere inside and along a simple closed contour $C$, then $\oint_C f(z) dz = 0$. - The Residue Theorem states that if $f(z)$ is meromorphic inside a contour $C$, then $\oint_C f(z) dz = 2\pi i \sum (\text{Residues at poles inside } C)$. - The sum of all residues of a rational function in the extended complex plane (including the point at infinity) is zero: $\sum \text{Res}(z_i) + \text{Res}(\infty) = 0$.

Step 1: Evaluate the integral $I_1$ along contour $C_1$.
The singularities (poles) of the function $f(z) = \frac{z}{(z-\alpha)(z-\beta)(z-\gamma)}$ are located at the points $z = \alpha$, $z = \beta$, and $z = \gamma$. The contour $C_1$ is defined by the disk $|z| < r$, where $r = \text{Min}\{\alpha, \beta, \gamma\}$. This means that the distance from the origin to the nearest pole is $r$. Therefore, there are absolutely no poles located inside the region bounded by $C_1$. Since $f(z)$ is completely analytic within and on $C_1$, by Cauchy's Integral Theorem: \[ I_1 = \oint_{C_1} f(z) dz = 0 \]

Step 2: Analyze the contour $C_2$ for integral $I_2$.
The second contour is defined as $C_2 : |z| = R+1$, where $R = \text{Max}\{\alpha, \beta, \gamma\}$. Because the radius of this circular path is strictly greater than the maximum absolute value among all poles, all three poles ($z = \alpha, \beta, \gamma$) lie entirely inside $C_2$. By the Residue Theorem: \[ I_2 = 2\pi i \left[ \text{Res}(f, \alpha) + \text{Res}(f, \beta) + \text{Res}(f, \gamma) \right] \]

Step 3: Compute the sum of residues using the residue at infinity.
Instead of calculating three separate algebraic residues, we can use the macro property of complex functions: the sum of all residues at finite poles is equal to negative the residue of the function at infinity: \[ \sum \text{Res}(f, \text{finite}) = \text{Res}\left[ - \frac{1}{w^2} f\left(\frac{1}{w}\right), 0 \right] \] Let us substitute $z = \frac{1}{w}$ into $f(z)$: \[ f\left(\frac{1}{w}\right) = \frac{\frac{1}{w}}{\left(\frac{1}{w} - \alpha\right)\left(\frac{1}{w} - \beta\right)\left(\frac{1}{w} - \gamma\right)} = \frac{\frac{1}{w}}{\frac{(1-\alpha w)(1-\beta w)(1-\gamma w)}{w^3}} = \frac{w^2}{(1-\alpha w)(1-\beta w)(1-\gamma w)} \] Now, apply the transformation formula for the residue at infinity: \[ - \frac{1}{w^2} f\left(\frac{1}{w}\right) = - \frac{1}{w^2} \cdot \frac{w^2}{(1-\alpha w)(1-\beta w)(1-\gamma w)} = \frac{-1}{(1-\alpha w)(1-\beta w)(1-\gamma w)} \] Evaluating this expression at $w = 0$: \[ \text{Res}(f, \infty) = \frac{-1}{(1-0)(1-0)(1-0)} = -1 \] Since the residue at infinity is $-1$, the sum of all finite residues inside $C_2$ must be: \[ \sum \text{Res}(finite) = -\text{Res}(f, \infty) = -(-1) = 1 \] Wait, let's verify via direct asymptotic behavior. As $z \to \infty$, $f(z) \sim \frac{z}{z^3} = \frac{1}{z^2}$. Since $f(z)$ decays as $\frac{1}{z^2}$, the residue at infinity is defined as $\lim_{z \to \infty} -z f(z) = \lim_{z \to \infty} -\frac{1}{z} = 0$. Let's double-check the calculation: \[ f(z) = \frac{z}{z^3 (1-\alpha/z)(1-\beta/z)(1-\gamma/z)} = \frac{1}{z^2} (1 + \dots) \] Ah! The residue at infinity is the coefficient of $\frac{1}{z}$ with a negative sign in the Laurent expansion around infinity. Since there is no $\frac{1}{z}$ term (the leading power is $\frac{1}{z^2}$), the residue at infinity is indeed $0$. Therefore: \[ \sum \text{Res}(finite) = 0 \quad \Rightarrow \quad I_2 = 2\pi i (0) = 0 \]

Step 4: Sum the two integrals.
Combining the results from Step 1 and
Step 3: \[ I_1 + I_2 = 0 + 0 = 0 \] This matches Option (B).