Finding the Smallest Positive Integer \(n\) for a Complex Equation

The question asks for the smallest positive integer \(n\) for which the complex expression \({\left( {\frac{{1 + i}}{{1 - i}}} \right)^n}\) equals 1. To solve this, we first need to simplify the complex number inside the parenthesis, \(\frac{{1 + i}}{{1 - i}}\).

Simplifying the Complex Fraction

To simplify a complex fraction of the form \(\frac{a+bi}{c+di}\), we multiply both the numerator and the denominator by the conjugate of the denominator. The denominator here is \(1 - i\), and its conjugate is \(1 + i\).

Let's perform the multiplication:

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

Using the distributive property (or FOIL) for the numerator and the difference of squares formula \((a-b)(a+b) = a^2 - b^2\) for the denominator:

Numerator: \((1 + i)(1 + i) = 1 \cdot 1 + 1 \cdot i + i \cdot 1 + i \cdot i = 1 + i + i + i^2\) Since \(i^2 = -1\), the numerator becomes \(1 + 2i - 1 = 2i\).

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

So, the simplified fraction is:

\[ \frac{{2i}}{2} = i \]

Thus, the original equation becomes \((i)^n = 1\).

Finding the Smallest Positive Integer \(n\) for \(i^n = 1\)

We need to find the smallest positive integer \(n\) such that the \(n\)-th power of \(i\) is equal to 1. Let's examine the first few positive integer powers of \(i\):

We can see a pattern here. The powers of \(i\) repeat in a cycle of length 4: \(i, -1, -i, 1\). The value \(i^n\) is equal to 1 when \(n\) is a positive integer multiple of 4. That is, \(n\) must be in the set \(\{4, 8, 12, 16, \dots\}\).

We are looking for the smallest positive integer \(n\) in this set.

The smallest positive integer \(n\) for which \(i^n = 1\) is 4.

Conclusion

The smallest positive integer \(n\) for which \({\left( {\frac{{1 + i}}{{1 - i}}} \right)^n} = 1\) is 4.

Revision Table: Key Concepts

Additional Information: Properties of Complex Numbers

• A complex number is typically written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, with \(i^2 = -1\).
• The modulus of a complex number \(z = a+bi\) is \(|z| = \sqrt{a^2+b^2}\). The modulus of \(i\) is \(|i| = \sqrt{0^2+1^2} = 1\). If \(|z|=1\), then \(|z^n| = |z|^n = 1^n = 1\). This tells us the result will have magnitude 1, but not necessarily be 1.
• Complex numbers can be represented in polar form \(r(\cos \theta + i \sin \theta)\) or exponential form \(re^{i\theta}\). Here, \(i\) corresponds to \(1(\cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})) = e^{i\frac{\pi}{2}}\). Then \(i^n = (e^{i\frac{\pi}{2}})^n = e^{i\frac{n\pi}{2}}\). For \(i^n = 1\), we need \(e^{i\frac{n\pi}{2}} = 1\). This occurs when \(\frac{n\pi}{2}\) is a multiple of \(2\pi\). So, \(\frac{n\pi}{2} = 2k\pi\) for some integer \(k\). This simplifies to \(\frac{n}{2} = 2k\), or \(n = 4k\). For the smallest positive integer \(n\), we take \(k=1\), which gives \(n=4\).