Understanding the Complex Number Problem

The question asks for the smallest positive integer \(n\) such that the complex expression \( \left(\frac{1-i}{1+i}\right)^{n^2} \) equals 1. Here, \(i\) is the imaginary unit, defined as \( \sqrt{-1} \). To solve this, we first need to simplify the base of the exponent, which is the complex fraction \( \frac{1-i}{1+i} \).

Simplifying the Complex Fraction \(\frac{1-i}{1+i}\)

To simplify a complex fraction involving \(i\) in the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator. The denominator is \(1+i\), and its conjugate is \(1-i\).

So, we calculate:

\( \frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} \)

Using the difference of squares formula \( (a+b)(a-b) = a^2 - b^2 \) for the denominator and expanding the numerator \( (a-b)^2 = a^2 - 2ab + b^2 \):

Numerator: \( (1-i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i + (-1) = 1 - 2i - 1 = -2i \)

Denominator: \( (1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2 \)

So, the simplified fraction is:

\( \frac{-2i}{2} = -i \)

Evaluating Powers of -i

Now the original equation becomes \( (-i)^{n^2} = 1 \). We need to find the powers of \(-i\) that result in 1.

Let's list the first few positive integer powers of \(-i\):

• \( (-i)^1 = -i \)
• \( (-i)^2 = (-1)^2 i^2 = 1 \times (-1) = -1 \)
• \( (-i)^3 = (-i)^2 \times (-i) = -1 \times (-i) = i \)
• \( (-i)^4 = (-i)^3 \times (-i) = i \times (-i) = -i^2 = -(-1) = 1 \)
• \( (-i)^5 = (-i)^4 \times (-i) = 1 \times (-i) = -i \)

The powers of \(-i\) repeat in a cycle of 4: \(-i, -1, i, 1\). For \( (-i)^k \) to be equal to 1, the exponent \(k\) must be a positive multiple of 4. That is, \(k\) must be of the form \(4m\) for some positive integer \(m\).

Finding the Smallest Positive Integer n

From the equation \( (-i)^{n^2} = 1 \), we determined that the exponent \(n^2\) must be a positive multiple of 4.

So, \( n^2 = 4m \), where \(m\) is a positive integer.

We are looking for the smallest positive integer value of \(n\). Let's test positive integer values for \(n\):

• If \( n=1 \), \( n^2 = 1^2 = 1 \). 1 is not a multiple of 4.
• If \( n=2 \), \( n^2 = 2^2 = 4 \). 4 is a multiple of 4 (4 = 4 \(\times\) 1). This satisfies the condition.

Since we found a positive integer \(n=2\) that satisfies the condition, and we started checking from the smallest positive integer (1), the smallest such integer \(n\) is 2.

Let's verify for \(n=2\):

\( \left(\frac{1-i}{1+i}\right)^{2^2} = (-i)^4 = 1 \)

This is indeed true.

Conclusion

The smallest positive integer \(n\) for which \( \left(\frac{1-i}{1+i}\right)^{n^2}=1 \) is 2.

Revision Table: Complex Number Concepts

Additional Information: Complex Number Forms

Complex numbers can be represented in different forms.

• Algebraic Form: \(z = a + bi\), where \(a\) is the real part and \(b\) is the imaginary part.
• Polar Form: \(z = r(\cos \theta + i \sin \theta)\) or \(z = re^{i\theta}\), where \(r = |z| = \sqrt{a^2 + b^2}\) is the modulus and \(\theta = \arg(z)\) is the argument (angle).

The expression \(\frac{1-i}{1+i}\) could also be simplified using polar form. \(1-i\) has modulus \( \sqrt{1^2 + (-1)^2} = \sqrt{2} \) and argument \(-\frac{\pi}{4}\). \(1+i\) has modulus \( \sqrt{1^2 + 1^2} = \sqrt{2} \) and argument \(\frac{\pi}{4}\). The quotient \(\frac{z_1}{z_2}\) has modulus \(\frac{r_1}{r_2}\) and argument \(\theta_1 - \theta_2\). So, the modulus is \( \frac{\sqrt{2}}{\sqrt{2}} = 1 \) and the argument is \( -\frac{\pi}{4} - \frac{\pi}{4} = -\frac{2\pi}{4} = -\frac{\pi}{2} \). A complex number with modulus 1 and argument \(-\frac{\pi}{2}\) is \( 1 \times (\cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2})) = 1 \times (0 + i(-1)) = -i \), which matches the algebraic simplification.

Raising a complex number in polar form \(re^{i\theta}\) to a power \(k\) is done using De Moivre's Theorem: \( (re^{i\theta})^k = r^k e^{ik\theta} = r^k (\cos(k\theta) + i \sin(k\theta)) \). For \( -i \), which is \( 1 \times e^{-i\pi/2} \), \( (-i)^{n^2} = 1^{n^2} e^{i n^2 (-\pi/2)} = e^{-i n^2 \pi/2} \). For this to be 1, the argument must be a multiple of \(2\pi\). So, \( -\frac{n^2 \pi}{2} = 2\pi m \) for some integer \(m\). \( -n^2/2 = 2m \Rightarrow n^2 = -4m \). Since \(n\) is a positive integer, \(n^2\) is positive, so \(m\) must be a negative integer, say \(m' = -m \gt 0\). \( n^2 = 4m' \). This leads to the same condition that \(n^2\) must be a positive multiple of 4.