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 = $\frac{1}{2\pi \mathrm{i}}{\int }_{{\gamma }_{\mathrm{r}}}\frac{{\mathrm{z}}^2+1}{(\mathrm{z}-\mathrm{a})(\mathrm{z}-\mathrm{b})}\mathrm{d}\mathrm{z}$ 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 ${\gamma }_r(t)=r{e}^{it}$ 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.