# $z^{2} + iz = 1$ (complex quadratic equation)

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)..

$z^2 + iz -1 = 0$

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

Any help appreciated!

• real numbers are complex numbers. – Thoth Jan 12 '17 at 13:31
• I fixed the matjax for $\sqrt{5}$ in your question, but I would guess you meant $z = (-1 \pm \sqrt{5})/2$ rather than $z = -1 \pm \sqrt{5}/2$ ? – TastyRomeo Jan 12 '17 at 13:36
• Yes, I did. Thank you – tcodeb Jan 12 '17 at 13:39

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.
• Is there a way of simplifying $sqrt(i^2+4)$ ? – tcodeb Jan 12 '17 at 13:45
• @tcodeb. Whet is $i^2$ equal to ? – Claude Leibovici Jan 12 '17 at 13:47
$$z=\frac{-i\pm\sqrt{i^2+4}}{2}=\frac{-i\pm\sqrt{3}}{2}$$
because $a=1$, $b=i$,$c=-1$ and $i^2=-1$.