Question

Real part of $\frac{(1+i{)}^2}{3-i}=$

• A. $-1/5$
• B. $1/5$
• C. $1/10$
• D. $-1/10$

Answer 1

The correct option is A $-1/5$

Given $\frac{(1+i{)}^2}{3-i}$

$=\frac{2i}{3-i}$ [Since $(i+1{)}^2=2i$]

$=\frac{2i(3+i)}{3^2-i^2}$ [After rationalizing the denominator]

$=\frac{2i(3+i)}{10}$

$=\frac{(3i-1)}{5}$.

Now real part of the given complex number is $-\frac{1}{5}$.