Complex Numbers — Maths STD 11 Science — Question

$a+(a+b) i+b(-1)=2+i \ldots \ldots .\left(\because i^2=-1\right)$

(a – b) + (a + b)i = 2 + i Equating real and imaginary parts, we get a – b = 2 ……(i) a + b = 1 …….(ii) Adding equations (i) and (ii), we get 2a = 3

$\therefore a=\frac{3}{2}$

Substituting $a=\frac{3}{2}$ in (ii), we get

$\frac{3}{2}+b=1$

$\therefore b=1-\frac{3}{2}=\frac{-1}{2}$

$\therefore a=\frac{3}{2}$ and $b=\frac{-1}{2}$