COMPLEX ANALYISIS: If z1 = 2 + i, z2 = 3 - 2i....

I have one problem here:

If z1 = 2 + i, z2 = 3 - 2i and evaluate
(i) z1^3 - 3z1^2 + 4z1

I tried to solve it by using De Moivre's formula,am I right? cos (ntheta)+isin(ntheta)
and I obtained theta from z1=2+i by using Tan(theta)=1/2=26.5 If am wrong please assist.

Why not do it the old fashion way? Just evaluate it:

\(\displaystyle \L\begin{array}{rcl}
\left( {a + bi} \right)^3 & = & a^3 + 3a^2 \left( {bi} \right) + 3a\left( {bi} \right)^2 + \left( {bi} \right)^3 \\
& = & \left( {a^3 - 3ab^2 } \right) + i\left( {3a^2 - b^3 } \right) \\
\end{array}.\)

\(\displaystyle \L\begin{array}{rcl}
\left( {a + bi} \right)^2 & = & a^3 + 2a\left( {bi} \right) + \left( {bi} \right)^2 \\
& = & \left( {a^2 - b^2 } \right) + i\left( {2ab} \right) \\
\end{array}.\)