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According to some references, following inequality constraint in $w \in \mathbb C^2$ is not convex. \begin{equation} \frac{|c^H w|^2}{\|Aw\|_2^2+\|Bw\|_2^2} \geq r_k, \qquad (1) \end{equation}
where $A$ and $B$ are $2 \times 2$ matrices and $c$ is a $2$-vector

However, after some phase rotation, the absolute value of the complex number in the numerator can be written as

\begin{equation} |c^H w| = \mathbb{R}\{e^{-j \theta} c^H w \}, \text{ where } \theta = \arg\{c^H w\}. \qquad (2) \end{equation} Then, after some manipulation, eqn (1) can be expressed again in the form of \begin{equation} \|\Phi w + d\| \leq \mathbb{R}\{e^{-j \theta} c^H w \} \qquad (3) \end{equation} which becomes a convex constraint as a SOCP (second order cone programming) form.

Here, I am very wondering why eqn (1) is a non-convex constraint, and eqn (3) is a convex constraint (How can we prove their non-convexity and convexity of (1) and (3), respectively ?).

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  • $\begingroup$ Is your question about what makes a constraint convex, or how to deal with complex numbers in constraints? $\endgroup$ – LarrySnyder610 May 3 at 20:57
  • $\begingroup$ What vector norm are you using? $\endgroup$ – Rodrigo de Azevedo May 4 at 11:57
  • $\begingroup$ (To LarrySnyder610) The constraint is a non-convex constraint. I am wondering why it is a non-convex, and what can make the constraint convex. \\ (To Rodirigo de Azevedo) The vector norm is 2-norm. $\endgroup$ – WNoh May 5 at 9:26
  • $\begingroup$ For a fix $\theta$, (3) is indeed convex. However, if you don't know $\theta$ and it is arbitrary, (3) is non-convex. $\endgroup$ – The Pheromone Kid May 13 at 10:48
  • $\begingroup$ $\theta$ is fixed. Why is (1) non-convex? How we could prove it? $\endgroup$ – WNoh May 13 at 10:51

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