a) Finding a root.

In order to find which one of the four choices is a root you have to evaluate $f(z)$ for each choice.If you find for a choice of $z$ that $f(z)=0$ then that choice of $z$ is a root of the equation.

Note that $$\begin{eqnarray*} \simplify{f({z1})} &=&\simplify[std]{{z1}^3+{-2*a1 -c1}*{z1} ^ 2 + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z1} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\ &=&\simplify[std]{{z1^3}+{-2*a1 -c1}{z1 ^ 2} + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * {z1} -{c1 * (a1 ^ 2 + b1 ^ 2)}}\\ &=&\simplify[std]{{z1^3}+{(( -2) * a1 -c1)*z1^2}+ {(2 * a1 * c1 + a1 ^ 2 + b1 ^ 2)*z1}-{c1 * (a1 ^ 2 + b1 ^ 2)}}\\ &=&0 \end{eqnarray*}.$$ So of the list of choices $z_1=\var{z1}$ is a root.

b) The other roots

Now that you have found a complex root it is very simple to find another complex root.

Since $f(z)$ is a polynomial with real coefficients then if $z=z_0$ is a root we have that the conjugate $z=\overline{z_0}$ is also a root.

Hence the complex number $z_2=\overline{\var{z1}}=\var{conj(z1)}$ is a root.

To find the real root $z_3=c$ we note that the constant term of $$f(z) =(z-z_1)(z-z_2)(z-c)$$ is $-z_1z_2c = -(\var{z1})(\var{conj(z1)})c=\var{-z1*conj(z1)}c$.

But we know that the constant term of $f(z)$ is $\simplify{-{c1 * (a1 ^ 2 + b1 ^ 2)}}$.

Hence $$\begin{eqnarray*} \var{-z1*conj(z1)}c &=&\simplify{-{c1 * (a1 ^ 2 + b1 ^ 2)}}\\ \Rightarrow c &=&  \simplify[]{{c1 * (a1 ^ 2 + b1 ^ 2)}/{abs(z1^2)}}\\ &=&\var{c1} \end{eqnarray*}$$