If z_1 = 2 - 3i and the roots of the equation | vedclass

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(C) Given the roots are $z_1 = 2 - 3i$,$z_2 = i$,and $z_3 = \bar{z}_1 = 2 + 3i$. The cubic equation is given by $(z - z_1)(z - z_2)(z - z_3) = 0$. Sub

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$9 - 10i$

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(C) Given the roots are $z_1 = 2 - 3i$,$z_2 = i$,and $z_3 = \bar{z}_1 = 2 + 3i$. The cubic equation is given by $(z - z_1)(z - z_2)(z - z_3) = 0$. Substituting the values: $(z - i)(z - (2 - 3i))(z - (2 + 3i)) = 0$. First,simplify the product of the last two factors: $(z - (2 - 3i))(z - (2 + 3i)) = ((z - 2) + 3i)((z - 2) - 3i) = (z - 2)^2 - (3i)^2 = z^2 - 4z + 4 + 9 = z^2 - 4z + 13$. Now,multiply by $(z - i)$: $(z - i)(z^2 - 4z + 13) = z^3 - 4z^2 + 13z - iz^2 + 4iz - 13i = z^3 + (-4 - i)z^2 + (13 + 4i)z - 13i = 0$. Comparing this with $z^3 + bz^2 + cz + d = 0$,we get $b = -4 - i$,$c = 13 + 4i$,and $d = -13i$. Therefore,$b + c + d = (-4 - i) + (13 + 4i) + (-13i) = (-4 + 13) + (-i + 4i - 13i) = 9 - 10i$.

If $z_1 = 2 - 3i$ and the roots of the equation $z^3 + bz^2 + cz + d = 0$ are $i$,$z_1$,and $\bar{z}_1$,then $b + c + d =$

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