Where $a$ is introduced in $a + bi$ when solving polynomials using complex numbers Trying to wrap my head around complex numbers and almost there. I am looking for problems that show me how to introduce $i$ into an equation. What I'm finding a lot of is "Simplify 2i + 3i = (2 + 3)i = 5i", where the $i$ has already been introduced somehow magically. The only primitive examples I've tried so far is "Simplify $\sqrt{-9}$" and by the definition of $i = \sqrt{-1}$ we get $3i$. That part makes sense for now.
But it's just $3i$, or $bi$ from the equation, there is no $a$. I don't see in what situations you get the $a$ and how you know how/where to add it. For example, on Wikipedia they show:

In this case the solutions are −1 + 3i and −1 − 3i, as can be verified using the fact that i2 = −1:
${\displaystyle ((-1+3i)+1)^{2}=(3i)^{2}=\left(3^{2}\right)\left(i^{2}\right)=9(-1)=-9,}$
${\displaystyle ((-1-3i)+1)^{2}=(-3i)^{2}=(-3)^{2}\left(i^{2}\right)=9(-1)=-9.}$

I am not skilled enough yet to know how they solved this, but I am wondering if they are saying $−1 + 3i$ is the form $a + bi$, or that $-1$ is separate.
Wondering if one could start off with a simple polynomial equation without any presence of $i$, and then show how you introduce $i$ in two different cases/examples:


*

*Where it's just $bi$, not $a + bi$

*Where it's $a + bi$
That way it should help explain how to introduce $i$ into a polynomial equation.
I'm imagining something like, or something more complicated if this doesn't have the $a$:
$(x + 3)^2 = -10$
I've started by doing:
$x + 3 = \sqrt{-10} = \sqrt{10}i$
$x = -3 + \sqrt{10}i$
Not sure if this means that $-3 + \sqrt{10}i$ is the complex number, or just $\sqrt{10}i$. Not sure if you need to be adding complex numbers to both sides, etc.

A complex number is always of the form $a+bi$, where $a,b\in \mathbb R$ but $a$ and/or $b$ can be zero. Also, when you were solving the equation $(x+3)^2=-10$ you forgot the second solution $x=-3-\sqrt{10} i$. Remember that $-10=(\pm \sqrt{10}i)^2$
