Question

The modulus of \((\frac{1+i}{1-i})^{75}-(\frac{1-i}{1+i})^{75}\)is

• A. 1
• B. 2
• C. \(\frac{1}{2}\)
• D. 4
• E. 16

Answer 1

The Correct Option is B

We need to find the modulus of the expression \[ \left( \frac{1 + i}{1 - i} \right)^{75} - \left( \frac{1 - i}{1 + i} \right)^{75}. \] First, simplify each of the terms \( \frac{1 + i}{1 - i} \) and \( \frac{1 - i}{1 + i} \). Using the formula for the modulus of a complex number \( \left| \frac{a + bi}{c + di} \right| = \frac{|a + bi|}{|c + di|} \), we can find that the modulus of both expressions is 1. Thus, \[ \left| \left( \frac{1 + i}{1 - i} \right)^{75} \right| = 1^{75} = 1, \quad \left| \left( \frac{1 - i}{1 + i} \right)^{75} \right| = 1^{75} = 1. \] The modulus of the difference is therefore \[ \left| 1 - 1 \right| = 2. \]

The correct option is (B) : \(2\)

Answer 2

Let's simplify the expressions inside the parentheses first:

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

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

So, the expression becomes: \(i^{75} - (-i)^{75}\)

Since 75 = 4 * 18 + 3, we have \(i^{75} = i^{4*18 + 3} = (i^4)^{18} * i^3 = (1)^{18} * i^3 = i^3 = -i\)

\((-i)^{75} = (-1)^{75} * i^{75} = -1 * i^{75} = -(-i) = i\)

Therefore, the expression simplifies to: \(-i - (i) = -2i\)

Now, we need to find the modulus of -2i:

\(|-2i| = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2\)

Therefore, the modulus of the given expression is 2.