方程 $z^4-4z^3+az^2+bz+78=0$ 的一个根为 $3+2i$，其中 $a$ 和 $b$ 为实数。求 $a$ 和 $b$ 的值。由此求方程的其他根，并给出其精确形式。

1. State the conjugate root and find the corresponding real quadratic factor.

Since the coefficients of the polynomial are real, the complex conjugate $3-2i$ is also a root.
The quadratic factor corresponding to the roots $3+2i$ and $3-2i$ is:
$(z - (3+2i))(z - (3-2i)) = z^2 - (3+2i + 3-2i)z + (3+2i)(3-2i)$
$= z^2 - 6z + (3^2 + 2^2) = z^2 - 6z + 13$

2. Express the quartic equation as a product of two quadratic factors and compare coefficients to find the second factor.

Let the second quadratic factor be $z^2 + pz + q$.
$z^4 - 4z^3 + a z^2 + bz + 78 = (z^2 - 6z + 13)(z^2 + pz + q)$
Comparing the constant terms:
$13q = 78 \implies q = 6$
Comparing the coefficients of $z^3$:
$p - 6 = -4 \implies p = 2$
So the second factor is $z^2 + 2z + 6$.

3. Compare coefficients to find the values of $a$ and $b$.

Comparing the coefficients of $z^2$:
$a = q - 6p + 13 = 6 - 6(2) + 13 = 7$
Comparing the coefficients of $z$:
$b = 13p - 6q = 13(2) - 6(6) = 26 - 36 = -10$

4. Solve the second quadratic factor to find the remaining roots.

To find the other roots, solve $z^2 + 2z + 6 = 0$ using the quadratic formula:
$z = \frac{-2 \pm \sqrt{2^2 - 4(1)(6)}}{2}$
$z = \frac{-2 \pm \sqrt{-20}}{2} = \frac{-2 \pm 2i\sqrt{5}}{2} = -1 \pm i\sqrt{5}$
The other roots of the equation are $3-2i$, $-1+i\sqrt{5}$, and $-1-i\sqrt{5}$.