1. We start with the given values: $x=1$ and $y=1$. 2. Calculate $z = \sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$. 3. Find the angle $\theta$ where $\tan(\theta) = \frac{y}{x} = \frac{1}{1} = 1$. 4. Therefore, $\theta = 45^\circ = \frac{\pi}{4}$ radians. 5. Write $z$ in trigonometric form: $$z = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right).$$ 6. To find the $n$th roots (here roots of order 3, 5, 6 are indicated), use De Moivre's theorem: $$W_k = \sqrt[n]{\sqrt{2}} \left( \cos \left( \frac{\pi}{4} + \frac{2k\pi}{n} \right) + i \sin \left( \frac{\pi}{4} + \frac{2k\pi}{n} \right) \right),$$ where $k=0,1,\dots,n-1$. 7. For example, for $n=3$ and $k=0$: $$W_0 = \sqrt[3]{\sqrt{2}} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right).$$ 8. Similarly, the document lists roots for $n=5$ and $n=6$ with varying $k$ values, adjusting angles accordingly. 9. These roots are complex numbers evenly spaced on the circle of radius $\sqrt[n]{\sqrt{2}}$ at angles shifted by $\frac{2k\pi}{n}$ starting from $\frac{\pi}{4}$. Final answer: $$z = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right),$$ with its $n$th roots given by $$W_k = \sqrt[n]{\sqrt{2}} \left( \cos \left( \frac{\pi}{4} + \frac{2k\pi}{n} \right) + i \sin \left( \frac{\pi}{4} + \frac{2k\pi}{n} \right) \right) \quad \text{for } k=0,1,\dots,n-1.$$