If $z_1=2-3 i$ and the roots of the equation $\mathrm{z}^3+\mathrm{b} \mathrm{z}^2+\mathrm{cz}+\mathrm{d}=0$…
If $z_1=2-3 i$ and the roots of the equation
$\mathrm{z}^3+\mathrm{b} \mathrm{z}^2+\mathrm{cz}+\mathrm{d}=0$ are $\mathrm{i}, \mathrm{z}_1$ and $\bar{z}_1$, then $\mathrm{b}+\mathrm{c}+\mathrm{d}=$
13
7
$9-10 \mathrm{i}$
$10-10 \mathrm{i}$
Solution
Let $z_1=2-3 i, z_2=i$ and $z_{3^{\prime}}=\bar{z}_1=2+3 i$ are the root of the equation
$\left(z-z_1\right)\left(z-z_2\right)\left(z-z_3\right)=0$
$\begin{aligned} & \Rightarrow \quad(z-i)(z-(2-3 i))(z-(2+3 i))=0 \\ & \Rightarrow \quad(z-i)\left(z^2-4 z+13\right)=0 \\ & \Rightarrow \quad z^3-4 z^2+13 z-z^2 i+4 z i-13 i=0 \\ & \Rightarrow \quad z^3+z^2(-4-i)+z(13+4 i)-13 i=0\end{aligned}$
The given equation is
$z^3+b z^2+c z+d=0$ ...(ii)
Comparing eqn. (i) with equation (ii), we get :-
$b=-4-i, c=13+4 i, d=0-13 i$
Then, $b+c+d=9-10 i$.