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: Understanding the Concept:
A complex number is real when its argument is a multiple of \(\pi\). Polar form makes powers easy.

Step 2: Polar Forms:
\(1+i=\sqrt2\,e^{i\pi/4}\) and \(1-i=\sqrt2\,e^{-i\pi/4}\).

Step 3: Simplify the Ratio:
\[ \frac{(1+i)^n}{(1-i)^{n-2}} = \frac{(\sqrt2)^n e^{in\pi/4}}{(\sqrt2)^{n-2}e^{-i(n-2)\pi/4}} = 2\,e^{i\pi(2n-2)/4} = 2\,e^{i\pi(n-1)/2} \]

Step 4: Reality Condition:
The number is real when \(\dfrac{\pi(n-1)}{2}\) is a multiple of \(\pi\), that is when \(n-1\) is even, so \(n\) is odd.
The smallest positive odd integer is \(n=1\). Check: for \(n=1\) the value is \(\dfrac{1+i}{(1-i)^{-1}}=(1+i)(1-i)=2\), which is real.

Final Answer:
The smallest such \(n\) is 1, option (A). \[ \boxed{\text{(A) } 1} \]