Question

If $\frac{a+3l}{2+ib}=1-1$, show that $(5a-7b)=0$

Answer 1

Simplify $\frac{a+3i}{2+ib}=1-i$

$a+3i=\left(1-i\right)\left(2+ib\right)$

$a+3i=\left(2+b\right)+i\left(b-2\right)$

By comparing both sides,

$a=2+b$

$3=b-2$

$b=5$

Put these values of $a$ and $b$ in left hand side of $5a-7b=0$.

$5\left(2+b\right)-7\left(5\right)$

$=5\left(2+\left(5\right)\right)-7\left(5\right)$

$=0$

$=RHS$