1. **State the problem:** We are given a complex number $z = x + it$ and the equation $\arg(z-2) - \arg(z+2) = \frac{\pi}{4}$. We need to find the Cartesian form of this equation. 2. **Recall the argument difference formula:** For two complex numbers $a$ and $b$, the difference of their arguments is the argument of their quotient: $$\arg(a) - \arg(b) = \arg\left(\frac{a}{b}\right)$$ 3. **Apply this to our problem:** $$\arg(z-2) - \arg(z+2) = \arg\left(\frac{z-2}{z+2}\right) = \frac{\pi}{4}$$ 4. **Express $z$ in terms of $x$ and $t$:** $$z = x + it$$ So, $$\frac{z-2}{z+2} = \frac{(x-2) + it}{(x+2) + it}$$ 5. **Find the argument of the quotient:** The argument of a complex number $\frac{A + iB}{C + iD}$ is $$\arg\left(\frac{A + iB}{C + iD}\right) = \arctan\left(\frac{B}{A}\right) - \arctan\left(\frac{D}{C}\right)$$ 6. **Calculate the argument difference:** $$\arg\left(\frac{z-2}{z+2}\right) = \arctan\left(\frac{t}{x-2}\right) - \arctan\left(\frac{t}{x+2}\right) = \frac{\pi}{4}$$ 7. **Use the tangent subtraction formula:** $$\tan\left(\arctan\left(\frac{t}{x-2}\right) - \arctan\left(\frac{t}{x+2}\right)\right) = \tan\left(\frac{\pi}{4}\right) = 1$$ 8. **Apply the formula for tangent of difference:** $$\frac{\frac{t}{x-2} - \frac{t}{x+2}}{1 + \frac{t}{x-2} \cdot \frac{t}{x+2}} = 1$$ 9. **Simplify numerator and denominator:** Numerator: $$\frac{t}{x-2} - \frac{t}{x+2} = t \left(\frac{1}{x-2} - \frac{1}{x+2}\right) = t \frac{(x+2) - (x-2)}{(x-2)(x+2)} = t \frac{4}{x^2 - 4}$$ Denominator: $$1 + \frac{t^2}{(x-2)(x+2)} = 1 + \frac{t^2}{x^2 - 4} = \frac{x^2 - 4 + t^2}{x^2 - 4}$$ 10. **Rewrite the equation:** $$\frac{t \frac{4}{x^2 - 4}}{\frac{x^2 - 4 + t^2}{x^2 - 4}} = 1$$ 11. **Simplify the complex fraction:** $$\frac{4t}{x^2 - 4} \cdot \frac{x^2 - 4}{x^2 - 4 + t^2} = 1$$ 12. **Cancel common factors:** $$\frac{\cancel{4t}}{\cancel{x^2 - 4 + t^2}} = 1$$ Actually, the $x^2 - 4$ cancels out, so we have: $$\frac{4t}{x^2 - 4 + t^2} = 1$$ 13. **Multiply both sides by denominator:** $$4t = x^2 - 4 + t^2$$ 14. **Rearrange to standard Cartesian form:** $$x^2 - 4 + t^2 - 4t = 0$$ 15. **Complete the square for $t$:** $$t^2 - 4t = (t^2 - 4t + 4) - 4 = (t - 2)^2 - 4$$ 16. **Substitute back:** $$x^2 - 4 + (t - 2)^2 - 4 = 0$$ 17. **Simplify constants:** $$x^2 + (t - 2)^2 - 8 = 0$$ 18. **Final Cartesian form:** $$x^2 + (t - 2)^2 = 8$$ **Answer:** The Cartesian form of the given equation is $$\boxed{x^2 + (t - 2)^2 = 8}$$