I have gotten to a chapter in my pre-calculus textbook where it mentions this equation: $|a+bi| = \sqrt{a^2+b^2}$
It doesn't really explain why that equation is true, and it just sort of moves on afterwards. This made no sense to me but normally when I can't figure something out I can just break it down and it just works because its math and math is consistent so that generally resolves the issue. In this case, it has just made me incredibly confused.
To my understanding, absolute value takes a number and leaves it unchanged if it is positive and turns it into a positive number if it is negative. It then stands to reason that the only $2$ possible options are that $|a|=a$ or $|a|=(-a)$. If I apply that logic to this equation, then I get this.
$$\sqrt{a^2+b^2}=a+bi \text{ or } \sqrt{a^2+b^2}=-(a+bi)\\\implies
\sqrt{a^2+b^2}=a+bi\text{ or }\sqrt{a^2+b^2}=-a-bi\\\implies\sqrt{a^2+b^2}-a=bi \text{ or } \sqrt{a^2+b^2}+a=-bi\\\implies{\sqrt{a^2+b^2}-a\over b}=i \text{ or } {\sqrt{a^2+b^2}+a\over-b}=i\implies \sqrt{a^2+b^2}-a=-(\sqrt{a^2+b^2}+a)\implies\sqrt{a^2+b^2}-a=-\sqrt{a^2+b^2}-a\implies 1 =-1\text{?}$$
$1$ obviously does not equal $-1$ so what is going on here? So far the answers I have gotten are "$i$ doesn't work like that because its a nonreal number" which makes no sense at all given that regardless of whether or not it is real, it is still a constant so it should be subject to things like the transitive law, and "that's not supposed to be absolute value its a different thing called modulus" but from googling it seems like those are the exact same in this context. The only options I can think of are
$\boldsymbol{1)}$ I have a fundamental misunderstanding of how absolute value works (I don't think it's this because I have seen absolute value used like this when dealing with quadratic equations)
$\boldsymbol{2)}$ There is some other issue with this math (I don't think it's this because I have gone over it more than I can count)
$\boldsymbol{3)}$ This equation is incorrect (I don't think it's this because both my pre-calculus teacher and math book disagree with this and there is literally an entire chapter dedicated to complex numbers and this equation)
Currently the option I am leaning towards is option #3 because I know that less advanced math curriculums will oversimplify things sometimes to the point of just being incorrect, but this is pre-calculus. I feel like I would have gotten past that sort of thing.
Best Answer
It looks like the issue is that you are seeing $|a + bi| = \sqrt{a^2 + b^2}$ as a theorem to be proved. But that isn't what it is. It's the definition of what "$|\cdots|$" means when applied to complex numbers.
You're taking a something that you know about the real number definition and trying to generalize it to complex numbers. This is a great way of thinking, but it looks like you're learning that the property, "either $|a| = a$ or $|a| = -a$," doesn't work out so well for complex numbers. But that just means that it might be better to try to generalize some other property.
Well, we can also think as the absolute value as the distance from the origin. When you picture complex numbers as points on the complex plane, this naturally gives rise to the definition $|a + ib| = \sqrt{a^2 + b^2}$, as Masd pointed out.
One way to see if a generalization is "good" is to see if it preserves nice properties of the original definition. Real number absolute values satisfy these useful identities: $$ |x + y| \leq |x| + |y|,\\ |xy| = |x||y|. $$ And you can prove that the complex number definition satisfies these too.