Finding the Smallest Positive Integer n for a Complex Equation

The problem asks for the smallest positive integer \(n\) that satisfies the equation \(\left(\dfrac{1+i}{1-i}\right)^n=1\). This involves simplifying a complex number expression and understanding the powers of \(i\).

Simplifying the Complex Fraction \(\left(\dfrac{1+i}{1-i}\right)\)

First, let's simplify the base of the expression, which is the complex fraction \(\dfrac{1+i}{1-i}\). We can do this by multiplying the numerator and the denominator by the conjugate of the denominator. The conjugate of \(1-i\) is \(1+i\).

$$ \dfrac{1+i}{1-i} = \dfrac{1+i}{1-i} \times \dfrac{1+i}{1+i} $$

Now, we expand the numerator and the denominator:

• 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:

$$ \dfrac{2i}{2} = i $$

The original equation now becomes \(i^n = 1\). We need to find the smallest positive integer n for which \(i^n\) equals 1. This requires knowledge of powers of complex numbers, specifically the powers of \(i\).

Understanding the Powers of \(i\)

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

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

The powers of \(i\) repeat in a cycle of four: \(i, -1, -i, 1\). The value \(i^n\) equals 1 when \(n\) is a positive multiple of 4. We are looking for the smallest positive integer n that makes \(i^n = 1\).

From the list of powers, the smallest positive integer \(n\) for which \(i^n = 1\) is 4. Any positive integer \(n\) of the form \(4k\) (where \(k\) is a positive integer) will satisfy \(i^n=1\), but the smallest such positive integer n is when \(k=1\), which gives \(n=4\).

Checking the Options for the Smallest Positive Integer n

The given options are 8, 12, 4, and 16. Let's see which of these positive integers is the smallest and satisfies the equation \(i^n = 1\):

• If \(n=8\), \(i^8 = (i^4)^2 = 1^2 = 1\). This is a solution, but not the smallest positive integer n.
• If \(n=12\), \(i^{12} = (i^4)^3 = 1^3 = 1\). This is a solution, but not the smallest positive integer n.
• If \(n=4\), \(i^4 = 1\). This is a solution.
• If \(n=16\), \(i^{16} = (i^4)^4 = 1^4 = 1\). This is a solution, but not the smallest positive integer n.

Comparing the positive integers that satisfy the equation, 4 is the smallest. Therefore, the smallest positive integer n is 4.

Understanding complex arithmetic and complex exponentiation is key to solving this type of equation solving problem. The pattern of powers of \(i\) provides a direct way to find the smallest positive integer n satisfying \(i^n=1\).

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