Let $a, b$ be real numbers, and there exists a complex number $z$, such that $|z| \leqslant 1$, and $z+\bar{z}|z|=a+b \mathrm{i}$. Then the maximum value of $a b$ is $\qquad$
$\qquad$.
7. $\frac{1}{8}$.

Let $z=r(\cos \theta+\mathrm{i} \sin \theta)$, then from the given conditions we have $a=r(1+r) \cos \theta, b=r(1-r) \sin \theta$, hence $a b=$ $r^{2}\left(1-r^{2}\right) \sin \theta \cos \theta \leqslant \frac{1}{2} r^{2}\left(1-r^{2}\right) \leqslant \frac{1}{2}\left(\frac{r^{2}+1-r^{2}}{2}\right)^{2}=\frac{1}{8}$. The equality holds when $r=\frac{\sqrt{2}}{2}, \theta=\frac{\pi}{4}$.