Question

The least positive integer n such that \(\frac{(2i)^n}{(1-i)^{n-2}}, i = \sqrt{-1}\), is a positive integer, is ___________.

Hint:

When dealing with powers of complex numbers, converting to polar form (\(re^{i\theta}\)) is very effective. For the result to be a positive real number, the argument (angle) of the resulting complex number must be an even multiple of \(\pi\) (i.e., \(2k\pi\)). Also, remember useful algebraic identities like \((1 \pm i)^2 = \pm 2i\).

Answer 1

Correct Answer: 6

Step 1: Understanding the Question 
We need to find the smallest positive integer \(n\) for which the given complex expression evaluates to a positive integer. 
Step 2: Key Formula or Approach 
We will simplify the complex numbers in the numerator and the denominator, preferably by converting them to their polar form \(re^{i\theta}\), which makes handling powers much easier. 
Step 3: Detailed Explanation 
Let the given expression be \(Z\). \[ Z = \frac{(2i)^n}{(1-i)^{n-2}} \] We convert the base of the numerator and the denominator to polar form.|
Numerator Base: \(2i\). The modulus is \(|2i| = 2\). The argument is \(\frac{\pi}{2}\). So, \(2i = 2e^{i\pi/2}\). 
Denominator Base: \(1-i\). The modulus is \(|1-i| = \sqrt{1^2+(-1)^2} = \sqrt{2}\). The argument is \(-\frac{\pi}{4}\). So, \(1-i = \sqrt{2}e^{-i\pi/4}\). Now, we substitute these polar forms back into the expression for \(Z\): \[ Z = \frac{(2e^{i\pi/2})^n}{(\sqrt{2}e^{-i\pi/4})^{n-2}} = \frac{2^n e^{i(n\pi/2)}}{(\sqrt{2})^{n-2} e^{-i(n-2)\pi/4}} \] Let's simplify the magnitude and the argument parts separately. 
Magnitude: \(|Z| = \frac{2^n}{(\sqrt{2})^{n-2}} = \frac{2^n}{2^{(n-2)/2}} = 2^{n - (n-2)/2} = 2^{(2n-n+2)/2} = 2^{(n+2)/2}\). 
Argument: \(\arg(Z) = \frac{n\pi}{2} - \left(-\frac{(n-2)\pi}{4}\right) = \frac{n\pi}{2} + \frac{n\pi}{4} - \frac{2\pi}{4} = \frac{2n\pi + n\pi - 2\pi}{4} = \frac{(3n-2)\pi}{4}\). So, the expression in polar form is \(Z = 2^{(n+2)/2} e^{i(3n-2)\pi/4}\). 
For \(Z\) to be a positive integer, three conditions must be met: The magnitude \(|Z| = 2^{(n+2)/2}\) must be an integer. This requires \((n+2)/2\) to be an integer, which means \(n+2\) must be an even number. This implies that \(n\) must be an even integer. The imaginary part of \(Z\) must be zero. This means the argument must be an integer multiple of \(\pi\). \[ \frac{(3n-2)\pi}{4} = k\pi \implies 3n-2 = 4k \] So, \(3n-2\) must be a multiple of 4. The real part must be positive. This requires the argument to be an even integer multiple of \(\pi\). \[ \frac{(3n-2)\pi}{4} = 2m\pi \implies 3n-2 = 8m \] So, \(3n-2\) must be a multiple of 8. We need to find the least positive even integer \(n\) such that \(3n-2\) is a multiple of 8. Let's test the smallest positive even integers: If \(n=2\): \(3(2)-2 = 4\). (Not a multiple of 8) If \(n=4\): \(3(4)-2 = 10\). (Not a multiple of 8) If \(n=6\): \(3(6)-2 = 16\). This is a multiple of 8 (since \(16 = 8 \times 2\)). The least positive integer \(n\) that satisfies all the conditions is 6. Let's verify for \(n=6\): \(Z = 2^{(6+2)/2} e^{i(3(6)-2)\pi/4} = 2^4 e^{i(16\pi/4)} = 16 e^{i(4\pi)} = 16(\cos(4\pi) + i\sin(4\pi)) = 16(1+0) = 16\). Since 16 is a positive integer, our answer is correct. 
Step 4: Final Answer 
The least positive integer n is 6.