If 1 – i is a root of the equation x

Solution:

for equation $x^{2}+a x+b=0$ one root is $1-i$.

Let us suppose other root is $p+i q$ when $p, q \in \mathbb{R}$

Sum of roots of quadratic equations is $=-a$

i.e. $(1-i)+(p+i q)=-a$

i. e. $1+p+i(-1+q)=-a$

since $a$ is real (given)

$\Rightarrow-1+q=0$                 (i. e. imaginary part is zero)

i.e. $q=1$

also product of roots $=b$

$\Rightarrow(1-i)(p+i q)=b$

i.e. $p-i^{2} q-i p+i q=b$

i. e. $p+q+i(q-p)=b$

Since $b$ is also real $\Rightarrow p-q=0$

$\Rightarrow p=q=1$

Hence $a=1+p=-2, b=p+q=1+1=2$