Question

You are given that 1+i is a root of the equation z^2+2z^2+az+b=0,where a and b are real numbers .which of the following is true?

Answer 1

$z^2+2z^2+az+b=0$

$\Rightarrow 3z^2+az+b=0$

As $1+i$ is a root of $3z^2+az+b=0$

So, $3(1+i)^2+a(1+i)+b=0$ [Put $z=1+i$ ]

$\Rightarrow$ $3(1+i^2+2i)+a+ia+b=0$

$\Rightarrow 3(1-1+2i)+a+ia+b=0$

$\Rightarrow 6i+ia+a+b=0$

$\Rightarrow i(6+a)+a+b=0$

Comparing real and imaginary parts both sides, we get

$6+a=0, a+b=0 \\\Rightarrow a=-6, b=-a \\\Rightarrow a=-6, b=6$