Question

$\text{If}(\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)=a+ib,\text{then}$ choose the correct option(s)

• A. $a=\frac{307}{442}$
• B. $b=\frac{599}{442}$
• C. $a=\frac{599}{442}$
• D. $b=-\frac{599}{442}$

Answer 1

Answer: (A), (B)

$b=\frac{599}{442}$

$\text{Given:}(\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)$

Taking L.C.M. and simplify
$=\left(\frac{(1+i)-2(1-4i)}{(1-4i)(1+i)}\right)\left(\frac{3-4i}{5+i}\right)$
$=\left(\frac{1+i-2+8i}{1+i-4i-4i^2}\right)\left(\frac{3-4i}{5+i}\right)[\because i^2=-1]$
$=\left(\frac{-1+9i}{5-3i}\right)\left(\frac{3-4i}{5+i}\right)$

Multiply two or more complex number
$=\frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}$
$=\frac{33+31i}{28-10i}$
$=\frac{33+31i}{2(14-5i)}$

On multiplying numerator and denominator by $(14+5i)$
$=\frac{(33+31i)}{2(14-5i)}\times \frac{14+5i}{14+5i}$
$=\frac{462+165i+434i+155(i{)}^2}{2\left(\right(14{)}^2-\left(5i{)}^2\right)}$
$=\frac{307+599i}{2(196-25i^2)}$
$=\frac{307+599i}{2(196+25)}$
$=\frac{307+599i}{442}$
$=\frac{307}{442}+\frac{599}{442}i$