$2.333 A=\left(\left(z^{2}+\frac{1}{z^{2}}\right)^{2}-4\left(z+\frac{1}{z}\right)^{2}+12\right)^{1 / 4}:(z-1)$.

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

$2.333 A=\left(\left(z^{2}+\frac{1}{z^{2}}\right)^{2}-4\left(z+\frac{1}{z}\right)^{2}+12\right)^{1 / 4}:(z-1)$.
Solution. Let $y = z + \frac{1}{z}$. Then $y^2 = z^2 + \frac{1}{z^2} + 2$, i.e., $z^2 + \frac{1}{z^2} = y^2 - 2$.

From this, $A = \frac{\sqrt[4]{(y^2 - 2)^2 - 4y^2 + 12}}{z - 1} = \frac{\sqrt[4]{y^4 - 8y + 16}}{z - 1} = \frac{\sqrt[4]{(y^2 - 4)^2}}{z - 1} = \frac{\sqrt{|y^2 - 4|}}{z - 1} = \frac{\sqrt{|z^2 + \frac{1}{z^2} + 2 - 4|}}{z - 1} = \frac{\sqrt{(z - \frac{1}{z})^2}}{z - 1} = \frac{|z - \frac{1}{z}|}{z - 1} = \frac{|z^2 - 1|}{(z - 1)|z|} = \frac{|z - 1||z + 1|}{|z|(z - 1)}$.

Consider 4 cases.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-022.jpg?height=147&width=510&top_left_y=79&top_left_x=117)

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-022.jpg?height=151&width=495&top_left_y=73&top_left_x=683)

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-022.jpg?height=151&width=510&top_left_y=226&top_left_x=115)
4) $\left\{\begin{array}{l}z > 1, \\ A = \frac{(z - 1)(z + 1)}{z(z - 1)}\end{array}\left\{\begin{array}{l}z > 1, \\ A = \frac{z + 1}{z}\end{array}\right.\right.$.

Answer: if $z \leq -1$ or $0 < z < 1$, then $A = \frac{z + 1}{z}$.