Let $z_{1} = a+ib$ and $z_{2} = c+id$. Thus, $|z_{1}|^2 = a^2+b^2, |z_{2}|^2 = c^2+d^2, Re(z_{1}z_{2}) =ac-bd$ and $Im(z_{1}z_{2}) =ad+dc$.
The problem can be posed as
\begin{equation}
\begin{array}{c}
minimize \hspace{1cm} a^2+b^2+ c^2+d^2 + ac-bd \\
s.t. \hspace{3cm} ad+bc = 1. \\
\end{array}
\end{equation}
Let $\mathbf{x} = [a \hspace{1mm} b \hspace{1mm} c \hspace{1mm} d]^{T}$. The previsous optimization problem can be rewrite as
\begin{equation}
\begin{array}{c}
minimize \hspace{1cm} \mathbf{x}^{T}A\mathbf{x} \\
s.t. \hspace{1cm} \mathbf{x}^{T}B\mathbf{x}= 1, \\
\end{array}
\end{equation}
where
\begin{equation}
A = \left [ \begin{array}{cccc}
1 & 0 & 1/2 & 0 \\
0 & 1 & 0 &-1/2 \\
1/2 & 0 & 1 & 0 \\
0 & -1/2 & 0 & 1 \\
\end{array} \right]
\end{equation} and
\begin{equation}
B = \left [ \begin{array}{cccc}
0 & 0 & 0 & 1/2 \\
0 & 0 & 1/2 &0 \\
0 & 1/2 & 0 & 0 \\
1/2 & 0 & 0 & 0 \\
\end{array} \right].
\end{equation}
Solving the problem for lagragian function $\mathcal{L} = \mathbf{x}^{T}A\mathbf{x} + \lambda (\mathbf{x}^{T}B\mathbf{x}-1)$, we have
$\frac{\partial \mathcal{L}}{\partial \mathbf{x}} = \mathbf{0} \Rightarrow (A+\lambda B)\mathbf{x} = \mathbf{0}$.
The linear system $(A+\lambda B)\mathbf{x} = \mathbf{0}$ has non-trivial solutions ($\mathbf{x} \neq \mathbf{0}$) only if $\lambda = \pm \sqrt{3}$ (The trivial solution $\mathbf{x} = \mathbf{0}$ does not satisfy the constraint $\mathbf{x}^{T}B\mathbf{x}=1$). In this case, $rank(A+\lambda B) = 2$.
For $\lambda = \pm \sqrt{3}$ , $\mathbf{x} = \alpha_{1}\mathbf{u}_{1} + \alpha_{2}\mathbf{u}_{2}$, where $\mathbf{u}_{1}$ and $\mathbf{u}_{2}$ are non-zero vectors in $Ker\{A+ \lambda B\}$, and $ \alpha_{1}$ and $ \alpha_{2}$ are any real parameters.
By substituting $\mathbf{x}$ in the previous optimization problem, we have the equivalent quadratic problem
\begin{equation}
\begin{array}{c}
minimize \hspace{1cm} \mathbf{\alpha}^{T}C\mathbf{\alpha} \\
s.t. \hspace{1cm} \mathbf{\alpha}^{T}D\mathbf{\alpha}= 1, \\
\end{array}
\end{equation} where $\mathbf{\alpha} = [\alpha_{1} \hspace{1mm} \alpha_{2}]^{T}$, and
\begin{equation}
C = \left [ \begin{array}{cc}
\mathbf{u}_{1}^{T}A\mathbf{u_{1}} & \mathbf{u}_{1}^{T}A\mathbf{u_{2}} \\
\mathbf{u}_{2}^{T}A\mathbf{u_{1}}& \mathbf{u}_{2}^{T}A\mathbf{u_{2}} \\
\end{array} \right],
\end{equation}
\begin{equation}
D = \left [ \begin{array}{cc}
\mathbf{u}_{1}^{T}B\mathbf{u_{1}} & \mathbf{u}_{1}^{T}B\mathbf{u_{2}} \\
\mathbf{u}_{2}^{T}B\mathbf{u_{1}}& \mathbf{u}_{2}^{T}B\mathbf{u_{2}} \\
\end{array} \right].
\end{equation}
For $\lambda = \sqrt{3}$, we can choose $\mathbf{u}_{1} = [ 1 \hspace{1.7mm} 0 \hspace{1.7mm} -1/2 \hspace{1.7mm} -\sqrt{3}/2]^{T}$, and
$\mathbf{u}_{2} = [ 0 \hspace{1.7mm} 1 \hspace{1.7mm} -\sqrt{3}/2 \hspace{1.7mm} 1/2]^{T}$. Thus,
\begin{equation}
C = \left [ \begin{array}{cc}
3/2 & 0 \\
0& 3/2\\
\end{array} \right],
\end{equation}
\begin{equation}
D = \left [ \begin{array}{cc}
-\sqrt{3}/2 & 0 \\
0& -\sqrt{3}/2\\
\end{array} \right].
\end{equation}
Here, there is no solution because the constraint becomes $\alpha_{1}^2+\alpha_{2}^2 = -2/\sqrt{3}$.
For $\lambda = -\sqrt{3}$, we can choose $\mathbf{u}_{1} = [ 1 \hspace{1.7mm} 0 \hspace{1.7mm} -1/2 \hspace{1.7mm} \sqrt{3}/2]^{T}$, and
$\mathbf{u}_{2} = [ 0 \hspace{1.7mm} 1 \hspace{1.7mm} \sqrt{3}/2 \hspace{1.7mm} 1/2]^{T}$. Thus,
\begin{equation}
C = \left [ \begin{array}{cc}
3/2 & 0 \\
0& 3/2\\
\end{array} \right],
\end{equation}
\begin{equation}
D = \left [ \begin{array}{cc}
\sqrt{3}/2 & 0 \\
0& \sqrt{3}/2\\
\end{array} \right].
\end{equation}
Here, the minimum value of the cost function is $\sqrt{3}$, for any $\alpha_{1}$ and $\alpha_{2}$ such that $ \alpha_{1}^2+\alpha_{2}^2 = 2/\sqrt{3}$.