Question

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

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Write 1+i and 1-i in polar form and compare the arguments.

Answer 1

The Correct Option is A

Step 1: Rationalise Differently:
Use $\dfrac{1+i}{1-i}=i$. Then
\[ \frac{(1+i)^n}{(1-i)^{n-2}} = \left(\frac{1+i}{1-i}\right)^{n-2}(1+i)^2 = i^{\,n-2}\cdot 2i = 2\,i^{\,n-1} \]

Step 2: Reality:
$i^{\,n-1}$ is real exactly when $n-1$ is even, because $i^0=1$, $i^2=-1$, while odd powers are $\pm i$.

Step 3: Answer:
The smallest positive $n$ with $n-1$ even is $n=1$, giving the value 2. Option (A).

Final Answer:
Option (A), n = 1. \[ \boxed{\text{(A) } 1} \]