One of the roots of the equation $z^4-4z^3+az^2+bz+78=0$, where $a$ and $b$ are real, is $3+2i$. Find the values of $a$ and $b$. Hence find the other roots of the equation, giving your answers in exact form.

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 $-1-i\sqrt{5}$.