# Convexity of an inequality constraint

According to some references, following inequality constraint in $$w \in \mathbb C^2$$ is not convex. $$$$\frac{|c^H w|^2}{\|Aw\|_2^2+\|Bw\|_2^2} \geq r_k, \qquad (1)$$$$
where $$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

$$$$|c^H w| = \mathbb{R}\{e^{-j \theta} c^H w \}, \text{ where } \theta = \arg\{c^H w\}. \qquad (2)$$$$ Then, after some manipulation, eqn (1) can be expressed again in the form of $$$$\|\Phi w + d\| \leq \mathbb{R}\{e^{-j \theta} c^H w \} \qquad (3)$$$$ 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 ?).

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