Norm estimate of a 2-by-2 matrix. Let $X\in M_2(\mathbb{C})$ with $\Vert\cos\theta X+\sin\theta X^*\Vert\leq 1$, for all $\theta\in[0,\frac{\pi}{2})$. Show that $\Vert \cos\theta e^{i\varphi}X+\sin\theta e^{-i\varphi}X^*\Vert\leq 1$, for all $\theta\in[0,\frac{\pi}{2})$ and $\varphi\in[0,2\pi).$

Comment: I could see it whenever $X$ is normal but could not prove it for general $X$.
Any comment is highly appreciated. Thanks in advance.
 A: As stated, the result is not true, even when $X$ is normal. Take $X=i\,I_2$. Then
$$
\|\cos\theta\,X+\sin\theta\,X^*\|=|\cos\theta-\sin\theta|\leq1,\ \qquad \theta\in\big[0,\tfrac\pi2\big);
$$
If you take $\varphi=\tfrac\pi2$, then
$$
\|\cos\theta\,e^{i\varphi}\,X+\sin\theta\,e^{-i\varphi}\,X^*\|
=|\cos\theta+\sin\theta|
$$
which can be as large as $\sqrt2$ when $\theta=\tfrac\pi4$.
Even if we assume the  condition working for all $\theta$, the inequality is still not true: let $$X=\tfrac{(1-i)}{\sqrt2}\,I_2.$$ Then for any $\theta$,
\begin{align}
\|\cos\theta\,X+\sin\theta\,X^*\|^2
&=\Big|\cos\theta\,\tfrac{(1-i)}{\sqrt2}+\sin\theta\,\tfrac{(1+i)}{\sqrt2}\Big|^2\\[0.3cm]
&=\tfrac12\,\Big|\cos\theta+\sin\theta-i(\cos\theta-\sin\theta)\Big|^2\\[0.3cm]
&=\tfrac12\,\Big((\cos\theta+\sin\theta)^2+(\cos\theta-\sin\theta)^2)\\[0.3cm]
&=1.
\end{align}
Now let $\theta=\varphi=\tfrac\pi4$. We have, for any $x,y$ with $\|x\|=\|y\|=1$,
\begin{align}
\langle (\cos\theta\,e^{i\varphi}\,X+\sin\theta\,e^{-i\varphi}\,X^*)x,y\rangle
&=\cos\theta\,e^{i\varphi}\,\tfrac{(1-i)}{\sqrt2}+\sin\theta\,e^{-i\varphi}\,\tfrac{(1+i)}{\sqrt2}\\[0.3cm]
&=\tfrac{\sqrt2}2\,\tfrac{(1+i)}{\sqrt2}\,\tfrac{(1-i)}{\sqrt2}+\tfrac{\sqrt2}2\,\tfrac{(1-i)}{\sqrt2}\,\tfrac{(1+i)}{\sqrt2}\\[0.3cm]
&=\sqrt2>1
\end{align}
