complex roots calulation question

Question

How can we find the roots of an equation such as:$z^2 +z +1=0 ,z \in \mathbb{C} $ ?

Answer 1

The discriminant of the quadratic equation: $az^2+bz+c=0,\; a,b,c\in\Bbb C,\; a\ne0$ is $$\Delta= b^2-4ac$$ and then their roots are $$z_1=\frac{-b+\sqrt{\Delta}}{2a} \quad;\quad z_2=\frac{-b-\sqrt{\Delta}}{2a}$$

Answer 2

Your equation is of the form $ax^2+bx+c=0$ therefore use the quadratic formula, $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ Therefore,$$z=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot1}}{2}$$ $$=\frac{-1\pm \sqrt{-3}}{2}.$$ Therefore, $$z=\frac{-1}{2}+\frac{\sqrt{3}}{2}i \ \ \ or \ \ \ z=\frac{-1}{2}-\frac{\sqrt{3}}{2}i.$$

Answer 3

I don't know why you don't want to solve it as a normal quadratic equation, but if you insist...

Let $z = re^{i\theta}$ (in polar form).

Then
$ z^2 + z + 1 = 0 \Rightarrow\\ r^2e^{2i\theta} + 2e^{i\theta} + 1 = 0 \Rightarrow\\ r^2(\cos 2\theta + i\sin 2\theta) + r(\cos \theta + i \sin \theta) + 1 = 0 \Rightarrow\\ r^2 \cos 2\theta + r \cos \theta + 1 = 0 \qquad \ldots (1)\\ r^2 \sin 2\theta + r \sin \theta = 0 \qquad \ldots (2) $

From $(2)$:
$ 2r^2 \sin \theta \cos \theta + r \sin \theta = 0 \Rightarrow\\ r\sin\theta(2r\cos\theta + 1) = 0 \Rightarrow\\ r\sin\theta = 0\ \text{or}\ 2r\cos\theta + 1 = 0 $

If $\sin \theta = 0$, then $\cos \theta = 1$ and $\cos 2\theta = 2\cos^2 \theta - 1 = 1$. Then from $(1)$, $r^2 + r + 1 = 0$, which is the original equation.

If $2r\cos\theta + 1 = 0$, then $\cos\theta = \dfrac{-1}{2r}$, and $\cos 2\theta = 2\cos^2 \theta - 1 = \dfrac{1}{2r^2} - 1$.

Then from $(1)$:
$r^2 \cos2\theta + r\cos\theta + 1 = 0 \Rightarrow\\ r^2\left(\dfrac{1}{2r^2} - 1\right) + r\left(\dfrac{-1}{2r}\right) + 1 = 0 \Rightarrow\\ \dfrac{1}{2} - r^2 - \dfrac{1}{2} + 1 = 0 \Rightarrow\\ r^2 = 1 \Rightarrow\\ r = 1$

We take the positive square root as $r = |z|$ is positive.

Thus, $\cos\theta = \dfrac{-1}{2r} = \dfrac{-1}{2}$ and $\sin\theta = \pm \dfrac{\sqrt 3}{2}$.

Therefore, the solutions are $\boxed{z = \dfrac{-1 \pm i \sqrt{3}}{2}}$.