4.1 4
Tom_Tiahrt 03 Mar 2009 14:54
Is this the general form of the solution to the problem?
(1)\begin{align} \sqrt{a + bi} &= \pm \left( \sqrt{\dfrac{a + \sqrt{a^2 + b^2}}{2}} - \ i \ \dfrac{b}{|b|} \sqrt{\dfrac{-a + \sqrt{a^2 + b^2}}{2}} \right) \end{align}
Then in this instance:
(2)\begin{align} z &= -\sqrt3 - i, \sqrt{z} = \pm \sqrt{-\sqrt3 - i}, \ a = (-\sqrt3 ), \ b = -1 \end{align}
Some of what I have seems to lead to the answer:
(3)\begin{align} z &= \pm \left( \sqrt{\dfrac{(-\sqrt3 ) + \sqrt{3 + 1}}{2}} - i \ \dfrac{-1}{|-1|} \ \sqrt{\dfrac{(-\sqrt3 ) - \sqrt{3 + 1}}{2}} \right) \\ &= \pm \left( \sqrt{\dfrac{(-\sqrt3 ) + 2}{2}} - i \ \dfrac{-1}{1} \ \sqrt{\dfrac{(-\sqrt3 ) - 2}{2}} \right) \\ &= \pm \left( \sqrt{\dfrac{(-\sqrt3 ) }{2} + \dfrac{2}{2}} - i \ (-1) \ \sqrt{\dfrac{(-\sqrt3 ) }{2} - \dfrac{2}{2} }\right) \\ &= \pm \left( \sqrt{\dfrac{(-\sqrt3 ) }{2} + 1} + i \ \sqrt{\dfrac{(-\sqrt3 ) }{2} - 1 }\right) \\ &= \pm \left( \sqrt{-0.8660254038 + 1} + i \ \sqrt{-0.8660254038 - 1 }\right) \\ &= \pm \left( \sqrt{0.1339745962} + i \ \sqrt{-1.8660254038}\right) \\ &= \pm \left( \sqrt{0.1339745962} + i \ \sqrt{(-1)(1.8660254038)}\right) \\ &= \pm \left( 0.3660254038 + (i \times i \times \sqrt{1.8660254038})\right) \\ &= \pm \left( 0.3660254038 + (i^2 \times \sqrt{1.8660254038})\right) \\ &= \pm \left( 0.3660254038 + (-1 \times 1.3660254038)\right) \\ &= \pm ( 0.3660254038 - 1.3660254038) \\ &= \pm \left( -1\right) \\ &= 1, -1 \end{align}
However, I do not understand where my arithmetic flaw is.