2A. Let the complex number $i$ be given. Determine all complex numbers $z$ such that the number $a=\frac{u-\bar{u} z}{1-z}$ is a real number.
Solution. First method. The number $a$ is real if and only if $a=\bar{a}$. In the problem $\frac{u-\bar{u} z}{1-z}=\overline{\left(\frac{u-\bar{u} z}{1-z}\right)}$, from which we get $\frac{u-\bar{u} z}{1-z}=\frac{\bar{u}-\bar{z}}{1-\bar{z}}$, and by simplifying the obtained equality, it follows that: $(\bar{u}-u)(1-z \bar{z})=0, z \neq 1$. From the last relation, we conclude that if $u=\bar{u}$, i.e., if $u$ is a real number, then the solution is any complex number $z \neq 1$. If $u$ is not a real number, then $1-z \bar{z}=0 \Leftrightarrow 1-|z|^{2}=0 \Leftrightarrow|z|=1$, i.e., the solution is any complex number whose modulus is equal to 1 ( $z \neq 1$ ).

Second method. Let $u=a+i b, z=x+i y$. Then

$$
a=\frac{u-\bar{u} z}{1-z}=\frac{a+i b-(a-i b)(x+i y)}{1-x-i y}=\frac{a-a x-b y+i(b-a y+b x)}{1-x-i y},
$$

and by rationalizing with the conjugate of the denominator of this fraction and simplifying the equations, we get $b\left(1-x^{2}-y^{2}\right)=0$, from which: $b=0$ or $x^{2}+y^{2}=1,(|z|=1)$. If $b=0$, i.e., if $u$ is a real number, then the solution is any complex number $z \neq 1$, and if $u$ is not a real number, then the solution is any complex number whose modulus is equal to 1 ( $z \neq 1$ ).