Question

Let $P\left(z\right)=z^3+a{z}^2+bz+c,$ where $a,b,c\in R$. If there exists a complex number $w$ such that the three roots of $P\left(z\right)$ are $w+3i,w+9i$ and $2w-4$, where $i^2=-1$, then the value of $a+b+c$ is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

• A. $136$
• B. $-136$
• C. $280$
• D. $-280$

Answer 1

The correct option is B $-136$

Let $w=x+iy$, so the three roots are
$x_1=x+i(y+3)x_2=x+i(y+9)x_3=(2x-4)+2iy$
Now,
$P\left(z\right)=z^3+a{z}^2+bz+c$
Sum of roots
$x_1+x_2+x_3=-a$
As $a\in R$, so
$Im(x_1+x_2+x_3)=0\Rightarrow 4y+12=0\Rightarrow y=-3$
Now,
$x_1=x{x}_2=x+6i{x}_3=(2x-4)-6i$
Product of roots
$x_1{x}_2{x}_3=-c$
As $c\in R$, so
$Im\left(x_1{x}_2{x}_3\right)=0\Rightarrow Im\left(x\right(x+6i\left)\right((2x-4)-6i\left)\right)=0\Rightarrow (2x-4)\times 6-6x=0\Rightarrow x=4$
Now,
$x_1=4x_2=4+6i{x}_3=4-6i$
Therefore,
$a=-12,c=-208,b=84$
Hence, $a+b+c=-136$