Question

The smallest positive (non-zero) integer "n" for which the expression \((\frac{1+i}{1-i})^n\) = 1 holds true, is ___.

Answer 1

Correct Answer: 4

To solve for the smallest positive integer \( n \) such that \(\left(\frac{1+i}{1-i}\right)^n=1\), we begin by determining the value of the expression \(\frac{1+i}{1-i}\).

First, rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator:

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

Calculate the numerator:

\((1+i)(1+i) = 1 + 2i + i^2 = 1 + 2i - 1 = 2i\) (since \(i^2 = -1\)).

Calculate the denominator:

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

Thus, \(\frac{1+i}{1-i} = \frac{2i}{2} = i\).

Now, we need \(i^n = 1\). The powers of \(i\) cycle every 4 terms as follows:

n	\(i^n\)
1	i
2	-1
3	-i
4	1

To achieve \(i^n = 1\), \(n\) must be a multiple of 4. The smallest positive integer satisfying this is \(n=4\).