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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?

StRichLee
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1 Answers1

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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.

Rondoudou
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  • Thanks for your answer,what I taught before was z=ρeiθ,ρ>0, θ∈(0,2π) in twelfth grade,and I just thought $\sqrt{x}$ is just to cut the degree half, so if in that case, their has no solution. Now I am wondering about is if I can combine two branches/determinations like take $\sqrt{2i}$ in θ∈(0,2π) and $\sqrt{-2i}$ in θ∈(-2π,0)or other branch just to make up the answer or it is fixed like only can use θ∈(−π,π) or θ∈(π,3π) when the branch was chosen? Sorry for replying so late though I have seen your answer immediately when you post but I take a lot of time to summarize and verify what I thought. – StRichLee Apr 20 '22 at 16:50
  • I think it depends on the context. Usually, you fix the square root once and for all on some subset of $\mathbb{C}$ and then you try to solve. – Rondoudou Apr 20 '22 at 17:21
  • I think the key point I stuck is $\sqrt{x}$ has two possible values in C,I finally figure out some way to pursuade myself XD,$x=i(-i)x$,so$\sqrt{x}=\sqrt{i(-i)x}$ and take θ∈(0,2π),if$x=ρe^{i(θ+2nπ)}$with a specific n'=k(like 0), and take x=2, θ of each square root i,(-i),x if we just cut the degree half is 1/4π,3/4π,0. The sum of degree will be 0=π which is weird, so $x=ρe^{i(θ+2nπ)}$ all of n is the solution of x at the same time, and we can't just let n be a specific k, but sometimes just fixed at one branch will take it easy to solve the problem. Thank you again for all your assistance. – StRichLee Apr 20 '22 at 18:21