Let a, b be two real numbers such that a < 0 < b. For a positive real number r, define γ_r(t) = re^it (where t ∈ |0, 2π|) and I_r = \(\rm \frac{1}{2\pi i}\int_{\gamma_r}\frac{z^2+1}{(z-a)(z-b)}dz\) Which of the following statements is necessarily true? 

Concept:

Residue Theorem:
 

The integral of a function around a closed contour in the complex plane is \( 2\pi i \) times the sum of the residues

of the function inside the contour. Hence, \(I_r \) will depend on whether \(a\) and /or \(b\) lie inside the contour defined by \(\gamma_r\) .

Explanation: The contour \( \gamma_r \) is a circle of radius \( \gamma_r \) centered at the origin, and the integrand has singularities

(poles) at \(z = a\) and \(z = b \) .

Poles of the integrand:

The function \( \frac{z^2 + 1}{(z - a)(z - b)} \) has two poles

at \( z = a\) and at \(z = b \)

\(I_r = \frac{1}{2\pi i} \int_{\gamma_r} \frac{z^2 + 1}{(z - a)(z - b)} \, dz\), where \(\gamma_r(t) = re^{it} \) for \(t \in [0, 2\pi]\) and \(a, b \) are real numbers such that \(a < 0 < b\) . 

The poles of the function \(\frac{z^2 + 1}{(z - a)(z - b)} \) within the contour \(\gamma_r\), which is a circle of radius  \(r\) centered at the origin.

The poles of the function are at \(z = a \) and \( z = b\) . Since \(a < 0\) and \(b > 0\) , the two poles are located on opposite sides of the origin.

Depending on the radius r , the contour \(\gamma_r\) may or may not enclose one or both of the poles.

Conditions for \( I_r \):

If the radius r is smaller than |a| (the absolute value of a ), then the contour does not enclose

any poles, so by the Cauchy integral theorem,  \( I_r \) = 0 .

If the radius r is greater than \(\max\{|a|, b\}\) , the contour encloses both poles, and by the residue theorem, \( I_r \) will be non-zero.

If r is between |a| and b , the contour may enclose exactly one pole, which will also make \( I_r \) non-zero.

Option 3: Ir = 0 if r > max {|a|, b} and |a| = b

This condition is true because if r > \(\max \{|a|, b\} \), the contour encloses both poles, leading to a situation

where the integral sums the residues at both poles, potentially canceling each other out.

Additionally, if \(|a| = b \), both poles are symmetrically placed, reinforcing the cancellation.

Thus, Option 3) is the correct answer.