The smallest positive integer $n$ for which $\dfrac{(1 + i)^n}{(1 - i)^{n-2}}$ is a real number, is ...

The given expression is $\dfrac{(1 + i)^n}{(1 - i)^{n-2}}$.

This can be rewritten as $\left(\dfrac{1 + i}{1 - i}\right)^n (1 - i)^2$.

Simplifying the term inside the bracket:

$\dfrac{1 + i}{1 - i} = \dfrac{(1 + i)^2}{(1 - i)(1 + i)} = \dfrac{1 + i^2 + 2i}{1^2 - i^2} = \dfrac{2i}{2} = i$

Simplifying the second term:

$(1 - i)^2 = 1 + i^2 - 2i = -2i$

Substituting these back into the expression, we get:

$i^n (-2i) = -2 i^{n+1}$

For this expression to be a real number, $i^{n+1}$ must be a real number. This is possible only when the exponent $n+1$ is an even integer.

$n + 1 = 2k$, where $k$ is an integer.

$n = 2k - 1$

To find the smallest positive integer $n$, substitute $k = 1$:

$n = 2(1) - 1 = 1$

Answer: $1$