Question

If $(\frac{1}{1+2i}+\frac{3}{1-i})\left(\frac{3-2i}{1+3i}\right)$ is reducible to $a+ib$, then values of $a$ and $b$ are

• A. $a=\frac{7}{10},b=\frac{-13}{5}$
• B. $a=\frac{7}{10},b=\frac{-11}{5}$
• C. $a=\frac{7}{11},b=\frac{-13}{10}$
• D. $a=\frac{7}{11},b=\frac{-13}{5}$

Answer 1

The correct option is B $a=\frac{7}{10},b=\frac{-11}{5}$

We have,
$(\frac{1}{1+2i}+\frac{3}{1-i})\left(\frac{3-2i}{1+3i}\right)=\left[\frac{(1-i)+(3+6i)}{(1+2i)(1-i)}\right]\left(\frac{3-2i}{1+3i}\right)=\frac{4+5i}{3+i}\times \frac{(3-2i)}{(1+3i)}=\frac{(12+10)+(15-8)i}{(3-3)+(9i+i)}=\frac{22+7i}{10i}=\frac{7}{10}+\frac{22}{10i}=\frac{7}{10}-\frac{11}{5}i(\because \frac{1}{i}=-i)\therefore a=\frac{7}{10},b=\frac{-11}{5}$