Question

Express in the standard form $\frac{(1+i{)}^2}{3-i}$

• A. $\frac{1}{5}-\frac{3}{5}i$
• B. $\frac{1}{5}+\frac{3}{5}i$
• C. $-\frac{1}{5}-\frac{3}{5}i$
• D. $-\frac{1}{5}+\frac{3}{5}i$

Answer 1

The correct option is D $-\frac{1}{5}+\frac{3}{5}i$

$\frac{(1+i{)}^2}{3-i}=\frac{1+2i+i^2}{3-i}$ .......... $(a+b{)}^2=a^2+2ab+b^2$

$=\frac{1+2i-1}{3-i}$ .......... $\because i=\sqrt{-1}⟹i^2=-1$

$=\frac{2i}{3-i}\times \frac{3+i}{3+i}$

$=\frac{6i+2i^2}{9-i^2}$

$=\frac{-2+6i}{10}$ .......... $\because i=\sqrt{-1}⟹i^2=-1$

$=-\frac{1}{5}+\frac{3}{5}i$

Ans: D