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

To solve for the least positive integer n such that \(\frac{(2i)^n}{(1-i)^{n-2}}\) is a positive integer, we start by analyzing the expression.

First, consider (2i)ⁿ:
(2i)ⁿ = 2ⁿiⁿ.

Now, consider (1-i)^n-2. We can express this using polar coordinates. The complex number \(1-i\) has a modulus \( \sqrt{1^2 + (-1)^2} = \sqrt{2} \) and angle \(-\frac{\pi}{4}\) (since it lies in the fourth quadrant). So in polar form, \(1-i = \sqrt{2}e^{-i\pi/4}\).

Thus, (1-i)^n-2 = (\sqrt{2}e^{-i\pi/4})^n-2:

• Magnitude: \( (\sqrt{2})^{n-2} = 2^{(n-2)/2} \)
• Angle: \( (n-2)(-\frac{\pi}{4}) = -\frac{\pi(n-2)}{4} \)

So (1-i)^n-2 = 2^{(n-2)/2}e^{-i\pi(n-2)/4}.

Now substitute back into the expression:

\(\frac{(2i)^n}{(1-i)^{n-2}} = \frac{2^ni^n}{2^{(n-2)/2}e^{-i\pi(n-2)/4}}\)

Simplify this:

\(=\frac{2^ni^n}{2^{(n-2)/2} \cdot e^{-i\pi(n-2)/4}} = 2^{n-(n-2)/2} i^n e^{i\pi(n-2)/4}\)

\(= 2^{n/2+1} i^n \cdot e^{i\pi(n-2)/4}\)

This will be a positive integer if:

1. The magnitude \(2^{n/2+1}\) is just an integer (it is always positive).
2. The imaginary unit \(i^n\cdot e^{i\pi(n-2)/4}\) results in the real part being positive while the imaginary part is zero.

For \(i^n\), recall the cycle: \(i^1=i\), \(i^2=-1\), \(i^3=-i\), \(i^4=1\), etc.

Find the combined argument of \(i^n\cdot e^{i\pi(n-2)/4}\):

• Argument of \(i^n: \frac{n\pi}{2}\)
• Argument of \(e^{i\pi(n-2)/4}: \frac{\pi(n-2)}{4}\)

Set total equal to \(2k\pi\) for integer \(k\), hence:

\(\frac{n\pi}{2} + \frac{\pi(n-2)}{4} = 2k\pi\)

Solve for \(n\):

\(\frac{2n + (n-2)}{4} = 2k\)

\((3n-2)/4 = 2k\)

\(3n-2 = 8k\)

\(3n = 8k + 2\)

\(n = \frac{8k+2}{3}\)

For divisibility, set \(8k+2 ≡ 0 \mod 3\):

Check \(8k ≡ -2 \mod 3 \rightarrow 2k ≡ 1 \mod 3\)

\(k ≡ 2 \mod 3\) implies the smallest positive is \(k=2\).

Then \(n = \frac{8(2)+2}{3} = \frac{16+2}{3} = 6\).

Thus, n = 6 is the smallest integer so that the expression is a positive integer, fitting within the given range (6,6).