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 the real part of a complex fraction, first simplify the denominator and then multiply by the conjugate.

Answer 1

The Correct Option is A

We need to find the real part of \[ \frac{1+2i}{(2-i)^2}. \] First simplify the denominator: \[ (2-i)^2=(2-i)(2-i). \] \[ =4-2i-2i+i^2. \] Since \[ i^2=-1, \] we get \[ (2-i)^2=4-4i-1. \] \[ =3-4i. \] So the expression becomes \[ \frac{1+2i}{3-4i}. \] To simplify, multiply numerator and denominator by the conjugate of the denominator. The conjugate of \[ 3-4i \] is \[ 3+4i. \] Thus, \[ \frac{1+2i}{3-4i} = \frac{(1+2i)(3+4i)}{(3-4i)(3+4i)}. \] Now simplify the numerator: \[ (1+2i)(3+4i) = 3+4i+6i+8i^2. \] \[ =3+10i+8(-1). \] \[ =3+10i-8. \] \[ =-5+10i. \] Now simplify the denominator: \[ (3-4i)(3+4i)=3^2+4^2. \] \[ =9+16. \] \[ =25. \] So, \[ \frac{1+2i}{(2-i)^2} = \frac{-5+10i}{25}. \] \[ = -\frac{1}{5}+\frac{2}{5}i. \] Therefore, the real part is \[ -\frac{1}{5}. \]