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

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.

Answer 1

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}$.

Answer 2

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 $2\pi$.