Complex Imaginations

Let the two roots that add up to $3 + 4i$ be $z_1 = p + qi$ and $z_3 = r + si$ . Then $p + r = 3$ and $q + s = 4$ .

By the complex conjugate root theorem , the other two roots must be $z_2 = p - qi$ and $z_4 = r - si$ . These have a product of $(p - qi)(r - si) = 13 + i$ , so $pr - qs = 13$ .

The coefficient of the $x^2$ term for roots $z_1$ , $z_2$ , $z_3$ , and $z_4$ is $b =$

$= z_1z_2 + z_1z_3 + z_1z_4 + z_2z_3 + z_2z_4 + z_3z_4$

$= (p + qi)(p - qi) + (p + qi)(r + si) + (p + qi)(r - si) + (p - qi)(r + si) + (p - qi)(r - si) + (r + si)(r - si)$

$= (p^2 + q^2) + (p + qi)(2r) + (p - qi)(2r) + (r^2 + s^2)$

$= p^2 + q^2 + 4pr + r^2 + s^2$

$= 2pr - 2qs + p^2 + 2pr + r^2 + q^2 + 2qs + s^2$

$= 2(pr - qs) + (p + r)^2 + (q + s)^2$

In this case, $pr - qs = 13$ , $p + r = 3$ , and $q + s = 4$ , so $b = 2(13) + 3^2 + 4^2 = \boxed{51}$ .

Let the 4 roots be $z_1,z_2,\overline{z_1},\overline{z_2}$

And let $z_1+z_2=3+4i,\overline{z_1}\overline{z_2}=13+i$

So, $\overline{z_1}+\overline{z_2}=3-4i,z_1z_2=13-i$

$b=z_1z_2+\overline{z_1}\overline{z_2}+(z_1+z_2)(\overline{z_1}+\overline{z_2})$

$=(13-i)+(13+i)+(3+4i)(3-4i)=51$

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Note: $\overline{z}$ is the complex conjugate of $z$

Let the four complex roots be $\alpha$ , $\beta$ , $\gamma$ , and $\delta$ . Let $\alpha + \beta = 3+4i$ and $\gamma \delta = 13+i$ . By Vieta's formula , we have $\alpha + \beta + \gamma + \delta = -a$ . Since $a$ is real, this means that $\gamma + \delta = 3-4i$ , the conjugate of $\alpha + \beta = 3+4i$ . Similarly, $\alpha \beta \gamma \delta = d$ , as $d$ is real, $\alpha \beta = \overline {\gamma \delta} = 13-i$ . Now we have:

$\begin{aligned} b & = \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta \\ & = 13 -i + (\alpha + \beta) (\gamma + \delta) + 13 +i \\ & = 26 + (3+4i)(3-4i) \\ & = 26 + 9 + 16 \\ & = \boxed{51} \end{aligned}$