Is there any algorithm to find complex eigen value of a matrix? I am solving eigen value equation for general n by n matrix and QR algorithm works pretty well for real eigen values and vectors for me. Is there an algorithm to find complex eigen values?
 A: Man, I think I understand your question.
Suppose you start with a matrix A. In case A has complex eigenvalues then the QR method will not give a triangular matrix as result (no matter how many iterations you make). However it will yield a Hessemberg matrix and you can deal with that fairly easy.
A: Here is an example using the standard algorithm. Start with
$$
 \mathbf{A} = 
\left[
\begin{array}{cr}
 3 & -2 \\
 4 & -1 \\
\end{array}
\right],
\tag{1}
$$
which has the eigenvalue spectrum
$$
 \lambda \left( \mathbf{A} \right) = \left\{ 
1 + 2 i, 1 - 2 i
\right\}.
$$
The decomposition is
$$
 \mathbf{A} = \mathbf{Q \, R} = 
% q
\frac{1}{5}
\left[
\begin{array}{rc}
 3 & 4 \\
 -4 & 3 \\
\end{array}
\right]
% r
\left[
\begin{array}{cr}
 5 & -2 \\
 0 & 1 \\
\end{array}
\right].
%
$$

Thanks to @littleO for the oversight. The eigenvalue problem does not care about the pedigree of the numeric field, $\mathbb{R}$ or $\mathbb{C}$. A demonstration:

The characteristic polynomial for a $2\times 2$ matrix is computed using the trace and determinant as
$$
 p \left( \lambda \right) 
= \det \left( \mathbf{A} - \lambda \mathbf{I}_{2} \right) 
= \lambda^{2} - \lambda \text{tr } \mathbf{A} + \det \mathbf{A}
$$
The intermediate values are 
$$
 \text{tr } \mathbf{A} = 2, \qquad \det \mathbf{A} = 5
$$
The characteristic polynomial is then
$$
  p \left( \lambda \right) = \lambda ^2-2 \lambda +5
$$
The eigenvalues are the roots of the characteristic polynomial
$$
  p \left( \lambda \right)  = 0 \qquad \Rightarrow \qquad \lambda \left( \mathbf{A} \right) = 
\left\{
  1 + 2 i, 1 - 2i
\right\},
$$
a conjugate pair.
