## Trigonometry (11th Edition) Clone

$x = 1.36+1.26~i$ $x = -1.775+0.546~i$ $x = 0.414-1.810~i$
$x^3+4-5i = 0$ $x^3 = -4+5i$ We can find $r$: $r = \sqrt{(-4)^2+(5)^2} = \sqrt{41} = 6.403$ We can find the magnitude of $x$: $r^{1/3} = \sqrt[3] {6.403} = 1.857$ Note that $~~-4+5i~~$ is in the second quadrant: $tan~\theta = -\frac{5}{4}$ $\theta = tan^{-1}(-\frac{5}{4})+180^{\circ}$ $\theta = 128.66^{\circ}$ Then: $\frac{128.65^{\circ}+360^{\circ}~k}{3} = 42.89^{\circ}+120^{\circ}~k$ We can find the solutions for $x$: When $k = 0$: $x = 1.857[cos~(42.89^{\circ})+i~sin~(42.89^{\circ})]$ $x = 1.36+1.26~i$ When $k=1$: $x = 1.857[cos~(162.89^{\circ})+i~sin~(162.89^{\circ})]$ $x = -1.775+0.546~i$ When $k=2$: $x = 1.857[cos~(282.89^{\circ})+i~sin~(282.89^{\circ})]$ $x = 0.414-1.810~i$