If I have two complex numbers $ a,b \in \mathbb{C} $，and $ |a|^2 + |b|^2 = 1 $，I need to get the max of \begin{aligned} (aa^*-bb^*)^2 - [(ab^*)^2 - (a^*b)^2]^2 \end{aligned} In the answer book, the auther let \begin{aligned} a = \cos \frac{\beta}{2} \\ b = \sin \frac{\beta}{2} e^{i\alpha} \end{aligned} then we have \begin{aligned} \cos ^2 \beta +\frac14 \sin ^4 \beta \sin ^2 2\alpha \end{aligned} In physics, $\beta$ is the angle with Z-ax, and $\alpha$ is the angle with X-ax.
But is this transformation is reasonably？ Why we can let $a$ with not phase factor? Does there are some math explain?
Thanks by LiZ.
$|a|^2+|b|^2=1\iff\begin{cases}|a|=\cos(t)\\|b|=\sin(t)\end{cases}\iff\begin{cases}a=\cos(t)e^{iu}\\b=\sin(t)e^{iv}\end{cases}\quad$ with $t,u,v$ reals.
Once developed the expression can be expressed as
$f(a,b)=\Big[\ 2|a|^4|b|^4+|a|^4+|b|^4-2|a|^2|b|^2\ \Big]-\underbrace{\Big(\bar a^4b^4+a^4\bar b^4\Big)}_g$
Since $|e^{iu}|=|e^{iv}|=1$ they are of no importance for the big square bracket.
On the other hand $g(a,b) = \cos(t)^4\sin(t)^4\Big(e^{i(4v-4u)}+e^{i(4u-4v)}\Big)$
But you can notice that if you take instead $\begin{cases}a'=\cos(t)\\b'=\sin(t)e^{i(v-u)}\end{cases}$
You'll get $\quad g(a,b)=g(a',b')$
Therefore in the end, WLOG you can set $\beta=2t$ and $\alpha=v-u$ and get the announced parametrisation.