Question

The real part of \[ \frac{1+2i}{(2-i)^2} \] is

• A. \(-\frac{1}{5}\)
• B. \(\frac{1}{5}\)
• C. \(-\frac{2}{5}\)
• D. \(\frac{2}{5}\)

Hint:

To find real part of a complex fraction, multiply numerator and denominator by the conjugate of the denominator.

Answer 1

The Correct Option is A

Step 1: First simplify denominator: \[ (2-i)^2=4-4i+i^2 \] \[ =4-4i-1=3-4i \]

Step 2: Therefore, \[ \frac{1+2i}{(2-i)^2} = \frac{1+2i}{3-4i} \]

Step 3: Multiply numerator and denominator by conjugate \(3+4i\): \[ \frac{1+2i}{3-4i}\cdot \frac{3+4i}{3+4i} \] \[ = \frac{(1+2i)(3+4i)}{3^2+4^2} \]

Step 4: Expand numerator: \[ (1+2i)(3+4i)=3+4i+6i+8i^2 \] \[ =3+10i-8=-5+10i \]

Step 5: \[ \frac{-5+10i}{25}=-\frac{1}{5}+\frac{2}{5}i \] So real part is: \[ \boxed{-\frac{1}{5}} \]