Let $z_1,z_2$ be complex numbers such that $Im(z_1z_2)=1$ Find the minimum value of $|z_1|^2+|z_2|^2+Re(z_1z_2)$ Question : 
Let $z_1,z_2$ be  complex numbers such that $Im(z_1z_2)=1$ Find the minimum value of  $|z_1|^2+|z_2|^2+Re(z_1z_2)$ 
I know that $|z_1+z_1| \leq |z_1|+|z_2|$ 
Also if I consider two complex numbers say $z_1 =x+iy$ and $z_2 =a+ib$ 
Now $|z_1|^2 =x^2+y^2 $ 
and $|z_2|^2 =a^2+b^2$
Also using A.M.-GM. inequality we have 
$\frac{x^2+y^2}{2}\geq \sqrt{2}xy$
$\Rightarrow x^2+y^2 \geq xy 2\sqrt{2}$ 
Also $a^2+y^2 =2\sqrt{2}ab$ 
Please suggest whether we can proceed in this manner also guide further on this thanks.
 A: 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}$.
A: Let $z_1 = r_1 e^{i\theta}$, $z_2 = r_2 e^{i\varphi}$, for $r_1, r_2 \ge 0$.  Then $\Im(z_1 z_2) = r_1 r_2 \sin (\theta + \varphi) = 1$, and we wish to minimize $$|z_1|^2 + |z_2|^2 + \Re(z_1 z_2) = r_1^2 + r_2^2 + r_1 r_2 \cos(\theta + \varphi)$$ subject to the above constraint.  Without loss of generality, $r_1 \ge r_2 > 0$; then for any $\alpha = \theta + \varphi$ satisfying $\sin \alpha = (r_1 r_2)^{-1}$ and $\cos \alpha > 0$, there exists $\alpha'$ such that $\sin \alpha' = \sin \alpha$ but $\cos \alpha < 0$, namely $\alpha' = \pi - \alpha$.  So we can write the function to be minimized as $$f(r_1, r_2) = r_1^2 + r_2^2 - r_1 r_2 \sqrt{1 - (r_1 r_2)^{-2}} = r_1^2 + r_2^2 - \sqrt{(r_1 r_2)^2 - 1},$$ where we clearly also require $r_1 r_2 \ge 1$.  A variety of methods can be used to minimize this function; one approach is to use the AM-GM inequality to get $$f(r_1, r_2) \ge 2 r_1 r_2 - \sqrt{(r_1 r_2)^2 - 1},$$ where upon noting that this is a function of the product $\rho = r_1 r_2$, has a minimum either at $\rho = 1$ or some critical point satisfying $$\frac{d}{d\rho} \left[ 2\rho - \sqrt{\rho^2 - 1} \right] = 0.$$  This latter condition gives the desired extremum $\rho = 2/\sqrt{3}$ corresponding to an attainable minimum $r_1 = r_2 = \rho^{1/2}$, $$f(r_1, r_2) = \sqrt{3},$$ for any $z_1, z_2$ whose angles add up to $\theta + \varphi = 2\pi/3$ modulo $2\pi$.
