B1. The equation is $6 z^{10}+a_{9} z^{9}+a_{8} z^{8}+a_{7} z^{7}+\ldots+a_{3} z^{3}+a_{2} z^{2}+a_{1} z+9216=0$, where $a_{1}, a_{2}, \ldots, a_{9}$ are real numbers and $z$ is the unknown. We know 5 solutions of the given equation, namely $1+i, 2+2 i, 3+3 i, 4+4 i$ and $\frac{1}{2}$, where $i$ is the imaginary unit. Determine the value of the coefficient $a_{9}$.
B1. If the polynomial on the left side of the equation has real coefficients, its complex roots appear in complex conjugate pairs. In addition to the given solutions of the equation, the solutions of the equation are also $1-i$, $2-2i$, $3-3i$, and $4-4i$. We thus know 9 roots of the polynomial. Since the polynomial is of degree 10, it has 10 roots counted with multiplicity. Let's denote its last root by $z_{0}$. The given equation is therefore equivalent to the equation

$$
\begin{aligned}
& 6\left(z-\frac{1}{2}\right)(z-(1+i))(z-(1-i))(z-(2+2 i))(z-(2-2 i)) \\
& \quad \cdot(z-(3+3 i))(z-(3-3 i))(z-(4+4 i))(z-(4-4 i))\left(z-z_{0}\right)=0
\end{aligned}
$$

If we multiply the left side of the equation, the constant term we get is the product

$$
6 \cdot \frac{1}{2} \cdot(1+i)(1-i)(2+2 i)(2-2 i)(3+3 i)(3-3 i)(4+4 i)(4-4 i) z_{0}=3 \cdot 2 \cdot 8 \cdot 18 \cdot 32 \cdot z_{0}
$$

where we have multiplied the complex numbers in conjugate pairs. Since the constant term in the equation must be equal to 9216, it follows that $z_{0}=\frac{9216}{3 \cdot 2 \cdot 8 \cdot 18 \cdot 32}=\frac{1}{3}$. To determine $a_{9}$, we need to consider what we get for the power $z^{9}$ when we multiply the left side of the above equation. The power $z^{9}$ is obtained when we take $z$ from 9 factors and a constant from one, so

$$
\begin{aligned}
a_{9} & =-6\left(\frac{1}{2}+(1+i)+(1-i)+(2+2 i)+(2-2 i)+(3+3 i)+(3-3 i)+(4+4 i)+(4-4 i)+z_{0}\right) \\
& =-6\left(\frac{1}{2}+20+\frac{1}{3}\right)=-125
\end{aligned}
$$

Note. The formulas for the constant term and the coefficient $a_{9}$ can also be obtained directly using Vieta's formulas.
Writing the other 4 roots or stating that the other 4 roots are the conjugate values. . . ..... 1 point
Writing the factorized form of the polynomial $z$ with the additional general root $\left(z-z_{0}\right)$. . . ..... 1 point
Calculating the last root $z_{0}$. . . ..... 2 points
Calculating $a_{9}$. ..... 3 points