Why does the positive root of a complex number output two solutions? I've pretty much just started the topic and didn't understand this point.
For $z=a+bi$
The root of $z$ is a complex number and can be expressed as $x+yi$
$ie. \sqrt{z}= x+yi$
$z=(x^{2}-y^{2})+(2xy)i$
Equating coefficients:
$a=x^{2}-y^{2}$ and $(2xy)i$
Solving simultaneously giving $x$ and $y$:
$x=\pm\sqrt{a+\sqrt{a^{2}+b^{2}}}, y=\frac{b}{2(\pm\sqrt{a+\sqrt{a^{2}+b^{2}}})}$
Essentially if $x_{+ve}=\sqrt{a+\sqrt{a^{2}+b^{2}}}$
Then $\sqrt{z} = \pm(x_{+ve}+iy_{+ve})$
This just seems wrong.
My understanding of the definition of a root was that the positive root of a number should only yield a positive answer.
$e.g. \sqrt{4} = 2$
I get behind the idea that $(\pm2)^{2}=4$ and I also get behind the idea that $(\pm(x_{+ve}+iy_{+ve}))^{2}=z$
Method two:
$z=|z|e^{i(\theta +2k\pi)}$ for $k\in\Bbb{Z}$
$\sqrt{z}=\sqrt{|z|}e^{\frac{i(\theta +2k\pi)}{2}}$
$\sqrt{z}=\sqrt{|z|}e^{i(\frac{\theta}{2}+\pi)}$ for $k=2n+1$ or $\sqrt{|z|}e^{i(\frac{\theta}{2})}$ for $k=2n, n\in\Bbb{Z}$
i.e. Two solutions
This method (if what I've done is valid), makes more sense than stating the positive root of a number has both a positive and a negative solution.
Am I missing something or is that the end of the story?
 A: I strongly urge you not to use the notation $\sqrt z$, unless $z$ is a non-negative real number.
Concerning the definition: a square root of a complex number $z$ is any complex number $w$ such that $w^2=z$. It turns out that every complex number other than $0$ has two distinct square roots ($0$ only has one square root, which is $0$ itself), and if $w$ is one of these roots, then the other one is $-w$. That's because if $w$ and $s$ are square roots of $z$, then $w^2=s^2=z$, but\begin{align}w^2=s^2&\iff w^2-s^2=0\\&\iff(w-s)(w+s)=0\\&\iff s=\pm w.\end{align}
So, you have tried to find a square root of $z=a+bi$ (with $a,b\in\Bbb R$) in two different ways you, in each approach, you got two distinct symmetric answers, which is what you should have got anyway.
A: In $\Bbb R$ there is an interaction between $<$ and the arithmetic operations $+$, $\times$ : For $a,b,c\in\Bbb R$ we have $a<b\implies a+c<b+c$ and we have $(a<b\land 0<c)\implies ac<bc$. It is not possible for this to work in $\Bbb C$ so we usually do not define any $<$ on $\Bbb C$ because it wouldn't have much (if anything) to do with the arithmetic.
Writing $\sqrt z$ when $z\in \Bbb C$ but $z\not\in \Bbb R^+\cup\{0\}$ should be avoided. There are two solutions to $x^2=z$ and it is ambiguous as to which one of them should be called $\sqrt z$.
