The small positive integer ' \(n\) ' for which \((1+i)^{2 n}=(1-i)^{2 n}\) is \(\begin{array}{lll}\text { (a) } 4 & \text { (b) } 8 & \text { (c) } 2\end{array}\) (d) 12

Let's first write the given expressions in polar form. To begin with, we'll find the angle and modulus for both complex numbers. For \(1+i\), the modulus is: \[r_1 = \sqrt{1^2 + 1^2} = \sqrt{2}\] The angle can be found as: \[\Theta_1 = \mathrm{arctan}\left(\frac{1}{1}\right) = \frac{\pi}{4}\] Similarly, for \(1-i\), the modulus is: \[r_2 = \sqrt{1^2 + (-1)^2} = \sqrt{2}\] The angle can be found as: \[\Theta_2 = \mathrm{arctan}\left(\frac{-1}{1}\right) = - \frac{\pi}{4}\] Now let's simplify \((1+i)^{2 n}\) and \((1-i)^{2 n}\). Using De Moivre's theorem, we have \[ (1+i)^{2n} = r_1^{2n} \times (cos(2n\Theta_1)+i\sin(2n\Theta_1)) \] \[ (1-i)^{2n} = r_1^{2n} \times (cos(2n\Theta_2)+i\sin(2n\Theta_2)) \] We are supposed to find n for which \((1+i)^{2 n}=(1-i)^{2 n}\). This means the real part and imaginary part of both expressions have to be equal.

Now let's equate the real parts of \((1+i)^{2 n}\) and \((1-i)^{2 n}\) as well as the imaginary parts. For the real part, considering that \(cos(-\theta) = cos(\theta)\), we have: \[ \cos(2n\Theta_1) = cos(2n\Theta_2) \] For the imaginary part, we have \(sin(-\theta) = -sin(\theta)\) giving us: \[ \sin(2n\Theta_1) = -\sin(2n\Theta_2) \] Since \(\Theta_1 = \Theta_2 =\frac{\pi}{4}\), then: \[ 2n\Theta_1 = 2n\Theta_2 \] The last equation has no solutions for small n.

Recall that the equality of complex numbers is satisfied if their angles differ by an integer multiple of \(2\pi\). Therefore, we can write the equation in terms of angles: \[ 2n\Theta_1 - 2n\Theta_2 = 2k\pi \] Substitute \(\Theta_1 = \frac{\pi}{4}\) and \(\Theta_2 = -\frac{\pi}{4}\). We are left with: \[ n = k \]

The smallest positive integer k (or n) that satisfies the problem is 1. However, by checking the given answer choice options, the smallest integer n available is 2. Therefore, the answer is: \[\boxed{n=2}\] (Option c)