how to find all the solutions to $z^2+z+1-i=0$?

Question

I have one exercise that I can't understand at all. My teacher isn't replying, and I need to solve it before the deadline..

The exercise says use the second line formula (ABC-formula) to find the complex solutions for this equation: $$z^2+z+1-i=0$$

Can someone explain how I can do this? I can't find anything on the internet that takes this topic, not even in the book.

I have already tried to read about it, but got absolutely nothing.

Answer 1

Hint: You can try with $z=a+bi$ where $a,b$ are real. Or you can solve it as quadratic equation.

If you multiply it by 4 we get: $$ 4z^2+4z+1+3-4i=0$$

so $$(2z+1)^2= -3+4i = (1+2i)^2$$

so $$2z+1 = \pm (1+2i)$$

Answer 2

Compute the discriminant as \begin{equation} \Delta = b^2 - 4ac = (1)^2 - 4(1)(1-i) = 1 - 4 + 4i = -3 + 4i \end{equation} Now write it in geometric form \begin{equation} \Delta = \vert \Delta \vert e^{i \theta} \end{equation} where \begin{equation} \Delta = \sqrt{(-3)^2 + 4^2} = 5 \end{equation} and \begin{equation} \theta = -\tan^{-1} \frac{4}{3} + \pi \end{equation} So \begin{equation} \Delta = 5e^{-i\tan^{-1} \frac{4}{3} + i\pi} = 1 + 2i \end{equation} The square root of this will be \begin{equation} z_1 = \frac{-b - \sqrt{\Delta}}{2a} \end{equation} and \begin{equation} z_2 = \frac{-b + \sqrt{\Delta}}{2a} \end{equation} i.e. \begin{equation} z_1 = \frac{-1 - 1 - 2i}{2} = -1 - i \end{equation} and \begin{equation} z_2 = \frac{-1 + 1 + 2i}{2} = i \end{equation}

Answer 3

The quadratic formula gives$$ z = \frac{1}{2}\left(-1 \pm \sqrt{1 - 4(1-i)}\right) = \frac{1}{2}\left(-1\pm\sqrt{-3+4i}\right)$$

The issue is dealing with the square root. Rewriting the complex number in exponential form is a common way forward. Remember that $a+ib = \sqrt{a^2+b^2}e^{i\theta}$ where $\theta$ is the polar angle specifying the point $a+ib$ on the unit circle. If you make sketch of $-3+4i$ and said angle, you ought to be able to solve for $$\theta = \arctan(-4/3) + \pi.$$ Then $$-3+4i = 5e^{i(\theta + 2\pi k)}$$ with $k$ integer to proceed. The extra $2\pi k$ is because the angle is not unique. We'll see that term doesn't matter for this problem, but in general, it can be important. Substituting this and raising to the $1/2$ power gives

$$ z = \frac{1}{2}\left(-1\pm\sqrt{5}e^{i(\theta/2 + \pi k)}\right)$$ $$ = \frac{1}{2}\left(-1\pm\sqrt{5}e^{i(\theta/2)}e^{i\pi k}\right)$$ $$ = \frac{1}{2}\left(-1\pm\sqrt{5}e^{i\theta/2)}\right)$$

In the last line, I drop the $e^{i\pi k} $ term, since it equals $1$ for $k$ even and $-1$ for $k$ odd. That is, no matter the $k$, we will still find two unique solutions. Now evaluate using Euler's formula.

$$ z = \frac{-1 \pm \sqrt{5}\left[\cos(\theta/2) + i\sin(\theta/2)\right]}{2} = -1-i, i $$

Answer 4

Alt. hint: $\;z^2+z+1-i=(z^2+1)+(z-i)=(z-i)(z+i)+(z-i)=(z-i)(z+1+i)\,$.