Let the two forces be $\vec{F}_1$ and $\vec{F}_2$, with magnitudes $F_1 = F_2 = F$ and directions $\vec{F}_1 = -\vec{F}_2$. Let the perpendicular distance between their lines of action be $l$.

Choose an arbitrary pivot point O. Let the perpendicular distance from O to the line of action of $\vec{F}_1$ be $d_1$, and to the line of action of $\vec{F}_2$ be $d_2$.

Case 1: The pivot O is located between the lines of action of the two forces. The two forces create torques in the same rotational direction. The total torque is the sum of the individual torques:

$$\tau = \tau_1 + \tau_2 = F d_1 + F d_2 = F(d_1 + d_2)$$

From the geometry, the sum of the perpendicular distances from the pivot to the lines of action is equal to the couple arm, $d_1 + d_2 = l$.

$$\tau = Fl$$

Case 2: The pivot O is located outside the lines of action of the two forces. The two forces create torques in opposite rotational directions. Assume the line of action of $\vec{F}_2$ is farther from O. The net torque is the difference between the individual torques:

$$\tau = \tau_2 - \tau_1 = F d_2 - F d_1 = F(d_2 - d_1)$$

From the geometry, the difference between the perpendicular distances from the pivot to the lines of action is equal to the couple arm, $d_2 - d_1 = l$.

$$\tau = Fl$$

In both cases, the torque is $\tau = Fl$. Since the location of the pivot O was arbitrary, the torque of the couple is independent of the pivot point and is equal to the product of the force magnitude and the couple arm.