Compute the adjoint map for complex number Consider the vector space $V = \mathbb{C}^2$ over $\mathbb{C}$ and the linear operator $φ: V \to V$
such that$φ(z_1, z_2)=(2z_1 + iz_2, (1 − i)z_1)$
Let $φ^*$ be the adjoint map of $φ$. Give the matrix representation of $φ^∗$ with respect to the standard basis of $(\mathbb{C}^2, \mathbb{C})$ and compute $φ^∗(u)$ for the vector
$$u =(3 − i, 1 + 2i)$$
For this question what I tried is the following 
I first get the matrix representation for $φ$ with respect to these matrices then I take the transpose of it since $φ^*=φ^T$. But I'm not sure if that's the correct way to solve it or either how to make the matrix representation for the complex number. What I did is to have a $(2\times4)$ matrix of each real and imaginary component of $z$ but I still think it's wrong. Can someone show me how to describe complex numbers in matrices?
 A: For complex vector spaces, the adjoint of a matrix is computed by transposing and complex conjugating all entries. For example
$\begin{pmatrix}1 & i \\ 2-i & i  \end{pmatrix}^* = \begin{pmatrix}1 & 2 + i \\ -i & -i  \end{pmatrix}$
A: You don't have to represent complex numbers as matrices since you are not considering them as group actions. Also, you need to write $\varphi^*$ with respect to the standard basis of $\mathbb{C}^2$, which is $\{(1,0),(0,1),(i,0),(0,i)\}$. So there is no need to write $i=\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}$.

Apparently, $\varphi$ could be represented by 
$$\begin{pmatrix}
i & 2\\
1-i & 0
\end{pmatrix}$$
and thus according to Richard Jensen's answer that

For complex vector spaces, the adjoint of a matrix is computed by transposing and complex conjugating all entries. 

The adjoint matrix, $\varphi^*$, of $\varphi$ is
$$\varphi^*=\begin{pmatrix}
-i  &1+i\\
2  &0
\end{pmatrix}\sim (z_1,z_2)\to((1+i)z_2-iz_1,2z_1)$$
Thus, $$\varphi^*(3−i,1+2i)=(-2,6-2i)$$
