1. Problem statement. Problem: $f(x) = x - \sqrt{x^2 - z}$. We will find the domain and a simplified algebraic form and discuss special cases. 2. Domain. The radicand must be nonnegative: $x^2 - z \ge 0$. Solve the inequality: $x^2 \ge z$. If $z \le 0$ then $x^2 \ge z$ holds for every real $x$, so the domain is all real numbers. If $z > 0$ then $x^2 \ge z$ means $|x| \ge \sqrt{z}$, so the domain is $x \le -\sqrt{z}$ or $x \ge \sqrt{z}$. 3. Simplify by rationalizing. Start with $f(x)=x-\sqrt{x^2 - z}$. Multiply numerator and denominator by the conjugate $x+\sqrt{x^2 - z}$ to get $$f(x) = \frac{(x-\sqrt{x^2 - z})(x+\sqrt{x^2 - z})}{x+\sqrt{x^2 - z}} = \frac{x^2 - (x^2 - z)}{x+\sqrt{x^2 - z}} = \frac{z}{x+\sqrt{x^2 - z}}.$$ 4. Validity and special case. The rationalized form $\frac{z}{x+\sqrt{x^2 - z}}$ is equivalent to the original wherever the denominator $x+\sqrt{x^2 - z}$ is nonzero. If $x+\sqrt{x^2 - z}=0$ then $\sqrt{x^2 - z}=-x$ and squaring yields $x^2 - z = x^2$ which forces $z=0$. Thus the denominator can vanish only when $z=0$ and $\sqrt{x^2}= -x$, which is true for $x \le 0$. For $z=0$ the original simplifies to $f(x)=x-\sqrt{x^2}=x-|x|$. Hence for $z=0$ we have $f(x)=0$ for $x\ge 0$ and $f(x)=2x$ for $x<0$. The rationalized formula fails at the points where the denominator is zero, but the original formula is defined there. 5. Final answer. Domain: If $z \le 0$ then domain is all real $x$. If $z > 0$ then domain is $x \in (-\infty,-\sqrt{z}] \cup [\sqrt{z},\infty)$. Simplified form: $$f(x)=\frac{z}{x+\sqrt{x^2 - z}}$$ valid when $x+\sqrt{x^2 - z} \neq 0$. Special case: For $z=0$ the function equals $f(x)=x-|x|$ which is $0$ for $x\ge 0$ and $2x$ for $x<0$.