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: 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 $-10=(\pm \sqrt{10}i)^2$
A: 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 $0+0i=0$.
A: When solving quadratic equations like $(x+3)^2 = -10$, then working out the square results in $x^2 + 6x + 19 = 0$. The discriminant of this equation is $D = 6^2 - 4 \times 19$, which is negative. This means that there are no real solutions, but instead two complex ones.
The (complex) solutions are $x=\frac{-6+\sqrt{D}}{2}$ and $x=\frac{-6-\sqrt{D}}{2}$. So as long as the discriminant is negative and the linear term (here $6x$) is non-zero, the solutions will be of the form $p + qi$.
A: If $x^2=-10$, then there are two different solutions, given by $x=\pm\sqrt{-10}=\pm i\sqrt{10}$. In other words, both $i\sqrt{10}$ and $-i\sqrt{10}$ are solutions. These would usually be called complex numbers (but usage varies), and since they have the form $bi$ (in the first $b=\sqrt{10}$ and in the second $b=-\sqrt{10}$), they might be called "(pure) imaginary" (but usage varies).
In your example, if $(x+3)^2=-10$, then $x+3=\pm\sqrt{-10}=\pm i\sqrt{10}$, so that $x=-3\pm i\sqrt{10}$. The two solutions are $-3+i\sqrt{10}$ and $-3-i\sqrt{10}$. They both have the form $a+bi$, where $a=-3$ for both and $b=\sqrt{10}$ for the first and $-\sqrt{10}$ for the second. They would not often be called imaginary, but everyone would call them complex numbers.
