Question

The real part of ${\left(\frac{1+i}{3-i}\right)}^2=$

• A. $1$
• B. $16$
• C. $16{\omega }^2$
• D. $\frac{-3}{25}$

Answer 1

The correct option is D $\frac{-3}{25}$

Consider
$z=\frac{1+i}{3-i}$
$=\frac{(1+i)(3+i)}{9+1}$
$=\frac{(3+i)(1+i)}{10}$
$=\frac{3+3i+i-1}{10}$
$=\frac{2+4i}{10}$
$=\frac{1+2i}{5}$
Now
$z^2=(\frac{1+2i}{5}{)}^2$
$=\frac{1-4+4i}{25}$
$=\frac{-3+4i}{25}$
Hence
$Re\left(z^2\right)=\frac{-3}{25}$