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.

You have a typo; the quadratic you considered has roots $-3\pm\sqrt{10}i$. But you're right about the result being a complex number.
It may help to think of it like this. Once we've invented $i$ and want to do arithmetic with it, we want to be able to add and multiply as much as we like while staying in our current number system. But you don't need an infinite-dimensional number system to accommodate every polynomial in $i$ with real coefficients, because $i^2+1=0$ ensures the general polynomial is always of the form $a+bi$ with $a,\,b$ real.
However, because we want addition and multiplication to keep us in the complex numbers, they must be defined to include the special case $a=0$ of imaginary numbers and the special case $b=0$ of real numbers, as well as their common member $0+0i=0$.
