Finding the least point of numerical range which lies on real axis Given $A\in\mathbb{C}^{n\times n}$, its numerical range is $W(A):=\{(\Re (x^*Ax),\Im(x^*Ax)) :x^*x=1\}\subseteq \mathbb{C}^2$.  $W(A)$ is a convex set.
An example of $W(A)$ plot is given in the figure below (red line is the boundary of $W(A)$).
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Given $A$, I want to find the point on the boundary of $W(A)$, the point which is minimum on the real-axis (blue point on the figure).

I think we need to minimize real-coordinate and make a constraint that imaginary-coordinate is zero:
\begin{array}{ll} \underset{x\in\mathbb{C}^{n}}{\text{min}} & \Re(x^*Ax)\\ \text{s.t.} & \Im(x^*Ax)=0,\\&x^*x=1.\end{array}
Since $\Re(x^*Ax)=\frac{A+A^*}{2}$ and $\Im(x^*Ax)=\frac{A-A^*}{2i}$, we get this problem
\begin{array}{ll} \underset{x\in\mathbb{C}^{n}}{\text{min}} & x^*(\frac{A+A^*}{2})x\qquad=&\underset{x\in\mathbb{C}^{n}}{\text{min}} & \frac{1}{2}(x^*Ax+x^*A^*x)\qquad=&\underset{x\in\mathbb{C}^{n}}{\text{min}} & x^*Ax\\ \text{ s.t.} & x^*(\frac{A-A^*}{2i})x=0,&\text{ s.t.} & x^*Ax=x^*A^*x,&\text{ s.t.} & x^*x=1\\&x^*x=1&&x^*x=1\end{array}
But $x^*Ax$ might be complex in general and minimization of $x^*Ax$ is not possible then, I couldn't find where I am doing wrong.
 A: I believe you cannot directly substitute the constraint into the optimization problem. This is not a complete answer but at least it reduces the search space to one dimension. Firstly we can convert the problem into a real one by defining $A=B+iC$ and $x=y+iz$ where entities of $B,C,y,z$ are real. Now
$$\begin{align*}
x^*Ax &= (y^T-iz^T)(B+iC)(y+iz)\\
&= \left[ y^TBy+z^TBz+z^T(C-C^T)y\right]+i\left[y^TCy+z^TCz+y^T(B-B^T)z\right]\\
&= \begin{bmatrix}y^T&z^T\end{bmatrix}\begin{bmatrix}B&0\\C-C^T&B\end{bmatrix}\begin{bmatrix}y\\z\end{bmatrix}+i\begin{bmatrix}y^T&z^T\end{bmatrix}\begin{bmatrix}C&B-B^T\\0&C\end{bmatrix}\begin{bmatrix}y\\z\end{bmatrix}
\end{align*}$$
which is now equivalent to the problem
$$\min_{u \in \mathbb{R}^n} u^T X u ~~~~ s.t.~~ u^T Y u = 0,~~ u^T u = 1$$
for real $X$ and $Y$. The Lagrange multipliers for this problem can be written as
$$ L := u^T X u - \alpha u^T Y u - \beta (u^T u - 1)$$
The solution must satisfy
$$\begin{align*}
\frac{\partial L}{\partial u} &= 0 \Rightarrow \left[(X+X^T)-\alpha(Y+Y^T)\right]u = 2 \beta u \\
\frac{\partial L}{\partial \alpha} &= 0 \Rightarrow u^T Y u = 0 \\
\frac{\partial L}{\partial \beta} &= 0 \Rightarrow u^T u = 1
\end{align*}$$
Now an algorithm to find a minimum would be calculating the eigenvectors of the matrix $(X+X^T)-\alpha(Y+Y^T)$ for varying $\alpha$ and checking if $u^T Y u=0$. Then the solution would be the minimum $\beta$ over all such eigenvectors.
