Let $q \in \mathbb{N}$, $q \geq 2$. Assume that for any $k \in \{1, \ldots, q\}$ we have \begin{align*} \sup_{z \in \mathbb{C}_+} |f_k(z)| = \sup_{w \in \mathbb{R}} |f_k(iw)|, \end{align*} where $\mathbb{C}_+ \stackrel{\text{def}}{=} \{z \in \mathbb{C} \ \colon \ \Re(z)>0\}$ denotes the open right half-plane. The functions $f_k$ are continuous and bounded on the closure of $\mathbb{C}_+$ and are analytic on $\mathbb{C}_+$.
I am looking for a way to show that \begin{align*} \sup_{z \in \mathbb{C}_+} \Big(\sum_{k=1}^q|f_k(z)|^2 \Big) = \sup_{w \in \mathbb{R}} \Big(\sum_{k=1}^q|f_k(iw)|^2\Big). \end{align*}
I would be grateful for any hint. I only seem to be able to show that \begin{align*} \sup_{z \in \mathbb{C}_+} \Big(\sum_{k=1}^q|f_k(z)|^2 \Big) \leq \sum_{k=1}^q \sup_{w \in \mathbb{R}} |f_k(iw)|^2, \end{align*} or \begin{align*} \sup_{z \in \mathbb{C}_+} \Big(\sum_{k=1}^q|f_k(z)|^2 \Big) \leq q \sup_{w \in \mathbb{R}} \Big(\sum_{k=1}^q|f_k(iw)|^2\Big). \end{align*} (using equivalence of the $\infty$-norm and the euclidean norm).