Question

Real part of $$\left( \frac{1+i}{1-i} \right) \left( \frac{2+i}{2-i} \right)$$ is:

• A. $\frac{3}{5}$
• B. $-\frac{3}{5}$
• C. $\frac{4}{5}$
• D. $-\frac{4}{5}$
• E. $-\frac{1}{5}$

Hint:

When simplifying expressions with complex numbers, remember to multiply by the conjugate of the denominator. This will eliminate the imaginary part from the denominator, making the expression easier to handle.

Answer 1

The Correct Option is D

We are given the expression: $$\left( \frac{1+i}{1-i} \right) \left( \frac{2+i}{2-i} \right)$$ First, we simplify each fraction by multiplying the numerator and denominator by the conjugate of the denominator.
Step 1: Simplifying $\frac{1+i}{1-i}$:
Multiply the numerator and denominator by $1+i$ (the conjugate of $1-i$): $$\frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}$$ The denominator becomes: $$(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2$$ The numerator becomes: $$(1+i)^2 = 1^2 + 2i + i^2 = 1 + 2i - 1 = 2i$$ Thus: $$\frac{1+i}{1-i} = \frac{2i}{2} = i$$ Step 2: Simplifying $\frac{2+i}{2-i}$:
Multiply the numerator and denominator by $2+i$ (the conjugate of $2-i$): $$\frac{2+i}{2-i} \times \frac{2+i}{2+i} = \frac{(2+i)^2}{(2-i)(2+i)}$$ The denominator becomes: $$(2-i)(2+i) = 2^2 - i^2 = 4 - (-1) = 5$$ The numerator becomes: $$(2+i)^2 = 2^2 + 2 \cdot 2 \cdot i + i^2 = 4 + 4i - 1 = 3 + 4i$$ Thus: $$\frac{2+i}{2-i} = \frac{3+4i}{5} = \frac{3}{5} + \frac{4i}{5}$$ Step 3: Now, multiply the two simplified expressions: $$\left( i \right) \left( \frac{3}{5} + \frac{4i}{5} \right)$$ Distribute $i$ across the terms: $$i \times \frac{3}{5} = \frac{3i}{5}, \quad i \times \frac{4i}{5} = \frac{4i^2}{5} = \frac{-4}{5}$$ Thus, the result is: $$\frac{3i}{5} - \frac{4}{5}$$
The real part of the complex number is $-\frac{4}{5}$.
Thus, the correct answer is option (D).