## Trigonometry (11th Edition) Clone

The two possible values of $x$ are: $cos~135^{\circ}+i~sin~135^{\circ}$ $cos~315^{\circ}+i~sin~315^{\circ}$
$x^2+i = 0$ $x^2 = -i$ $x = (-i)^{1/2}$ $x = [1(0-~i)]^{1/2}$ Let $z = 1(0-~i)$ $z = 1(cos~270^{\circ}+i~sin~270^{\circ})$ $r = 1$ and $\theta = 270^{\circ}$ We can use this equation to find the square roots: $z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$ When k = 0: $z^{1/2} = 1^{1/2}~[cos(\frac{270^{\circ}}{2}+\frac{(360^{\circ})(0)}{2})+i~sin(\frac{270^{\circ}}{2}+\frac{(360^{\circ})(0)}{2})]$ $z^{1/2} = cos~135^{\circ}+i~sin~135^{\circ}$ When k = 1: $z^{1/2} = 1^{1/2}~[cos(\frac{270^{\circ}}{2}+\frac{(360^{\circ})(1)}{2})+i~sin(\frac{270^{\circ}}{2}+\frac{(360^{\circ})(1)}{2})]$ $z^{1/2} = cos~315^{\circ}+i~sin~315^{\circ}$