Can you find the solutions of the equation:  \qquad qquad \qquad x^2 + i x - i \ = \ 0 \ "?"  Make sure to give your answers in standard complex form ( a + bi form).

Mar 3, 2018

$x = {17}^{\frac{1}{4}} / 2 \cos \alpha + i \left({17}^{\frac{1}{4}} / 2 \sin \alpha - 1\right)$ and ${17}^{\frac{1}{4}} / 2 \cos \alpha - i \left({17}^{\frac{1}{4}} / 2 \sin \alpha + 1\right)$, where $\alpha = {\tan}^{- 1} \left(\frac{2 \pm 2 \sqrt{17}}{4}\right)$

Explanation:

Solution of $a {x}^{2} + b x + c = 0$ is given by a quadratic formula as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

therefore for ${x}^{2} + i x - i = 0$

$x = \frac{- i \pm \sqrt{{i}^{2} + 4 i}}{2}$

= $\frac{- i \pm \sqrt{- 1 + 4 i}}{2}$

As $- 1 + 4 i = \sqrt{17} \left(\cos \theta + i \sin \theta\right)$, where $\tan \theta = - 4$

and using DeMoivre's theorem

$\sqrt{- 1 + 4 i} = {17}^{\frac{1}{4}} \left(\cos \alpha + i \sin \alpha\right)$, where $\theta = 2 \alpha$

As $\tan 2 \alpha = \frac{2 \tan \alpha}{1 - {\tan}^{2} \alpha} = - 4$

we have $4 {\tan}^{2} \alpha - 2 \tan \alpha - 4 = 0$ and

$\tan \alpha = \frac{2 \pm 2 \sqrt{17}}{4}$ and $\alpha = {\tan}^{- 1} \left(\frac{2 \pm 2 \sqrt{17}}{4}\right)$

Hence we have two roots of $- 1 + 4 i$ given by

${17}^{\frac{1}{4}} \left(\cos \alpha + i \sin \alpha\right)$, where $\alpha = {\tan}^{- 1} \left(\frac{2 \pm 2 \sqrt{17}}{4}\right)$

and $x = \frac{- i \pm {17}^{\frac{1}{4}} \left(\cos \alpha + i \sin \alpha\right)}{2}$

= ${17}^{\frac{1}{4}} / 2 \cos \alpha + i \left({17}^{\frac{1}{4}} / 2 \sin \alpha - 1\right)$

and ${17}^{\frac{1}{4}} / 2 \cos \alpha - i \left({17}^{\frac{1}{4}} / 2 \sin \alpha + 1\right)$,

where $\alpha = {\tan}^{- 1} \left(\frac{2 \pm 2 \sqrt{17}}{4}\right)$