$\sqrt{x}+\sqrt{-x}=2$. The answer is $x=-2i$ or $x=\pm2i$ For the prolem(1)
$$\sqrt{x}+\sqrt{-x}=2 \qquad (1)$$
I take $\sqrt{-x}$ as $\sqrt{x}\times i$.
In the end, I get
$$\sqrt{x}=1-i=\sqrt{2} e^{i( \frac{7}{4} \pi +2n\pi)}$$
$$x=-2i=2e^{i( \frac{7}{2} \pi + 4n\pi)}$$
But if I take $\sqrt{-x}$ as $\sqrt{x}\times (-i)$.The answer will become
$$\sqrt{x}=1+i=\sqrt{2}e^{i( \frac{1}{4} \pi +2n\pi)}$$ $$x=2i=2e^{i( \frac{1}{2} \pi + 4n\pi)}$$
So now the problem is I should take the principle branch $\sqrt{-1}=i$ or second branch $\sqrt{-1}=-i$ or even both of them$\sqrt{-1}=\pm i$ ? I have seen the problem Is $\sqrt{-1}$ equal to $i$ or $\pm i$ . And another question(2),
$$if \quad \sqrt{x}=1-i=\sqrt{2}e^{i( \frac{7}{4} \pi +2n\pi)},$$
$$\quad then \quad x=-2i=2e^{i( \frac{7}{2} \pi + 4n\pi)}\qquad \qquad(2)$$
$$or \quad x=-2i=2e^{i( \frac{7}{2} \pi + 2n\pi)}$$
Futher more, teacher taught us that $\sqrt{-1}=i$ and $z=p(\cos{\theta}+i \sin{\theta})$ when I was twelfth grade in Taiwan, so if I only use $\sqrt{-1}=i$ and $z=p(\cos{\theta}+i \sin{\theta})$ for the math problem for senior high school in Taiwan like the above$\sqrt{x}+\sqrt{-x}=2$, the answer is $x=-2i$ or what?
 A: The problem is that $\sqrt{x}$ has two possible values in $\mathbb{C}$ : if $z$ is such that $z^2 = x$, then $-z$ as well. You changed the sign in $\sqrt{-x}$ but, in $\mathbb{C}$, you could also decide to change the sign of $\sqrt{x}$.
A way to solve it through is to develop it until you have no more square roots, find solutions and see if they are indeed solutions of the initial equation. Here, you have :
$$ x - x + 2 \sqrt{-x^2} = 4 $$
and then :
$$ - 4 x^2 = 16 $$
Therefore, $x = \pm 2i$, and you know there can't be any more solutions. Then, coming back to your equation, you find that $\sqrt{2i} = 1 + i$ or $\sqrt{2i} = -1-i$ ; $\sqrt{-2i} = 1-i$ or $i-1$.
Usually, one choose the square root on $\mathbb{C} \setminus \mathbb{R}^{-}$ by writing $z = \rho e^{i \theta}$, $\rho > 0$, $\theta \in (- \pi, \pi)$, $\sqrt{z} := \sqrt{\rho} e^{i \theta / 2}$. In this case, $x = \pm 2i$ are both solutions.
However, if you choose another determination of the square root, you may find that it has either no solution at all or that both $2i$ and $-2i$ are solutions.
