How to show that following is a solution of the optimization problem?

Question

In one of the research paper there are two complex vectors $\mathbf{y}$ and $\mathbf{a}$ and there is a scalar $x$. It is shown that $$\|\mathbf{y}-\mathbf{a}\exp(jx)\|^2_2$$ is minimized over $x$ when $$\exp(-jx)=\frac{\mathbf{y}^\mathbb{H}\mathbf{a}}{|\mathbf{y}^\mathbb{H}\mathbf{a}|}.$$ I know that $$\|\mathbf{y}-\mathbf{a}\exp(jx)\|^2_2=\|\mathbf{y}\|^2+\|\mathbf{a}\|^2-2\mathrm{Re}\{\mathbf{ya}^\mathbb{H}\exp(jx)\}$$ and minimizing the above expression with respect to $x$ is equivalent to maximizing $$2\mathrm{Re}\{\mathbf{ya}^\mathbb{H}\exp(jx)\}$$ which is maximized when $x=-\arg(\mathbf{ya}^\mathbb{H})$. Now I do not know how this results in $$\exp(-jx)=\frac{\mathbf{y}^\mathbb{H}\mathbf{a}}{|\mathbf{y}^\mathbb{H}\mathbf{a}|}.$$ Any help in understanding this will be highly appreciated. Thanks in advance.

Answer 1

Assuming that $\cdot^\mathbb{H}$ denotes the transpose and complex conjugate, you actually have $$\|\mathbf{y}-\mathbf{a}\exp(jx)\|^2_2= \|\mathbf{y}\|^2+\|\mathbf{a}\|^2-2\mathrm{Re}\{\mathbf{y}^\mathbb{H}\mathbf{a}\exp(jx)\}.$$

Then this is minimized when $$ -x = \arg(\mathbf{y}^\mathbb{H}\mathbf{a}). $$

Then the claim follows using basic properties of complex numbers:

If we have $\exp(-jx)=\frac{\mathbf{y}^\mathbb{H}\mathbf{a}}{|\mathbf{y}^\mathbb{H}\mathbf{a}|}$, then, taking $\arg(\cdot)$, it follows that $$ \arg({\mathbf{y}^\mathbb{H}\mathbf{a}}) = \arg\left(\frac{\mathbf{y}^\mathbb{H}\mathbf{a}}{|\mathbf{y}^\mathbb{H}\mathbf{a}|}\right) \\ = \arg(\exp(-jx)) = -x + 2\pi k. $$ Without loss of generality we can consider $k=0$, so we get $ -x = \arg(\mathbf{y}^\mathbb{H}\mathbf{a}). $