Question

Solve the following equation:


`z {}^(4) + z {}^(2) - i √(3) = 0`
Note that:

`i = √( - 1)`
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Answer 1

Complete the square.

`z^4 + z^2 - i\sqrt 3 = \left(z^2 + \frac12\right)^2 - \frac14 - i\sqrt3 = 0`

`\left(z^2 + \frac12\right)^2 = \frac{1 + 4\sqrt3\,i}4`

Use de Moivre's theorem to compute the square roots of the right side.

`w = \frac{1 + 4\sqrt3\,i}4 = \frac74 \exp\left(i \tan^(-1)(4\sqrt3)\right)`

`\implies w^(1/2) = \pm \frac{\sqrt7}2 \exp\left(\frac i2 \tan^(-1)(4\sqrt3)\right) = \pm \frac{2+\sqrt3\,i}2`

Now, taking square roots on both sides, we have

`z^2 + \frac12 = \pm w^(1/2)`

`z^2 = \frac{1+\sqrt3\,i}2 \text{ or } z^2 = -\frac{3+\sqrt3\,i}2`

Use de Moivre's theorem again to take square roots on both sides.

`w_1 = \frac{1+\sqrt3\,i}2 = \exp\left(i\frac\pi3\right)`

`\implies z = {w_1}^(1/2) = \pm \exp\left(i\frac\pi6\right) = \boxed{\pm \frac{\sqrt3 + i}2}`

`w_2 = -\frac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \frac{5\pi}6\right)`

`\implies z = {w_2}^(1/2) = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i(5\pi)/(12)\right)}`