Solve this complex number: $z^2+(1+i)z+i=0$

Question

I need to solve:

$$z^2+(1+i)z+i=0$$

first of all I used $$z = (a+ib)$$ and I get:

$$(a+ib)^2+(1+i)(a+ib)+i=0$$ $$a^2-b^2+2aib+a+ib+ia-b+i=0$$

then I have ordered, on the left the number without the $i$ and on the right all the number with the $i$:

$$\underbrace{a^2-b^2+a-b}_\text{real} + \underbrace{2aib+ib+ia+i}_\text{imaginary} = 0$$

So I have set this:

$$\begin{cases} a^2-b^2+a-b=0 \\ 2aib+ib+ia+i \end{cases}$$

$$\begin{cases} a(a+1)-b(b-1) = 0 \\ 2aib+ib+ia+i \end{cases}$$

How can I procede? I can say that the first equation is true when $a=b$

Answer 1

hint: $z^2 + (1+i)z + i = 0 \Rightarrow z^2+z + iz+i = 0\Rightarrow z(z+1) + i(z+1) = 0 \Rightarrow ...$

Answer 2

Cool thing about complex polynomials is that they alwas have n solutions (where n is the order of the polynomial) and you can brute force it quite simply. $$z^2 + (1+i)z + i = 0 \implies$$ $$(z+\frac{1+i}{2})^2 = -i - (1+i)^2 = -3i = 3e^{\frac{-\pi}{2}i}$$

Now you can take a square root to get:

$$z +\frac{1+i}{2} = \sqrt{3}e^{\frac{-\pi}{2}i \pm \pi i}$$

Simplify and solve.

Answer 3

The others pointed out a few ways how to guess the solution or derive the result for this specific case. However, the beauty of complex numbers is also, that math works on them just fine. So... just use the quadratic formula

$$z=\frac{-(1+i)\pm \sqrt{(1+i)^2-4i}}{2}=\frac{-(1+i)\pm \sqrt{-2i}}{2}$$ $$=\frac{-(1+i)\pm (1-i)}{2}=\{-i,-1\}$$ I used $\sqrt{-2i}=\sqrt{2e^{-i\pi/2}}=\sqrt{2}(\cos(-\pi/4),\sin(-\pi/4))=1-i$.