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How to I solve this complex quadratic equation?

$z^2 + iz = 1$

What I did first was rearranging in the $ax^2 + bx + c = 0$ form then used the quadratic formula but I get real numbers as a solution (should be complex I believe)..

$z^2 + iz -1 = 0$

Quadratic formula gives:

$z = -1 \pm \sqrt{5}/2$

Any help appreciated!

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  • $\begingroup$ real numbers are complex numbers. $\endgroup$ – Thoth Jan 12 '17 at 13:31
  • $\begingroup$ I fixed the matjax for $\sqrt{5}$ in your question, but I would guess you meant $z = (-1 \pm \sqrt{5})/2$ rather than $z = -1 \pm \sqrt{5}/2$ ? $\endgroup$ – TastyRomeo Jan 12 '17 at 13:36
  • $\begingroup$ Yes, I did. Thank you $\endgroup$ – tcodeb Jan 12 '17 at 13:39
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Looks like you treated $b$ as $b = 1$, not $b = i$. The quadratic formula works over $\mathbb{C}$ as much as it works over $\mathbb{R}$ -- better, actually, since you can always take a square root.

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  • $\begingroup$ Is there a way of simplifying $sqrt(i^2+4)$ ? $\endgroup$ – tcodeb Jan 12 '17 at 13:45
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    $\begingroup$ @tcodeb. Whet is $i^2$ equal to ? $\endgroup$ – Claude Leibovici Jan 12 '17 at 13:47
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Using quadratic formula you get:

$$z=\frac{-i\pm\sqrt{i^2+4}}{2}=\frac{-i\pm\sqrt{3}}{2}$$

because $a=1$, $b=i$,$c=-1$ and $i^2=-1$.

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  • $\begingroup$ you are welcome! $\endgroup$ – Arnaldo Jan 12 '17 at 13:53

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