# Where is error in this solution to $\sqrt{x} + \sqrt{-x} = 2$?

Given the equation: $$\sqrt{x} + \sqrt{-x} = 2$$ The solutions are $$x = \pm 2i$$. This can be seen via Wolfram Alpha $$\left( \sqrt{x} + \sqrt{-x} \right)^2 = 2^2$$ $$\sqrt{x}^2 + 2\sqrt{x}\sqrt{-x} + \sqrt{-x}^2 = 4$$ $$x + 2\sqrt{x}\sqrt{-x} - x = 4$$ $$\sqrt{-x^2} = 2$$ $$-x^2 = 4$$ $$x = \pm 2i$$

However, my approach to the problem only found the positive value to this equation. $$\sqrt{x} + \sqrt{-x} = 2$$ $$\sqrt{x} + i\sqrt{x} = 2$$ $$(1+i) \sqrt{x} = 2$$ $$(1-i)(1+i) \sqrt{x} = 2(1-i)$$ $$2 \sqrt{x} = 2 (1-i)$$ $$x = (1-i)^2$$ $$x = 1^2 - 2i + (-i)^2 = -2i$$ Where did I make a mistake here that resulted in me only getting one of the two solutions to this equation?

• By pulling the $-1$ out from thr second square root you've already assumed a priori that $x > 0.$ Dec 31, 2022 at 20:49
• The error comes from the first deduction, where you assume $\sqrt{-1}=i$, whilst $-1$ has two roots in the complex plane (the other is $-i$). Dec 31, 2022 at 20:49
• I personally think that this question is erroneous to begin with because the domain of the square-root function should be non-negative only. Nevertheless, for your solution, if you let $y=-x$ then you will get $\sqrt{y} + i \sqrt{y} = 2$ as well. Dec 31, 2022 at 20:49
• $\sqrt{-x}=\pm i\sqrt{x}$, whence two sub-equations and two solutions, one for each sign. Dec 31, 2022 at 21:56

I will use $$z$$ instead of $$x$$, and assume that $$z \in \mathbb C$$. I will also use $$z^{1/2}$$ instead of $$\sqrt{x}$$ to denote the square root. Then your equation becomes $$f(z) = z^{1/2} + (-z)^{1/2} - 2 = 0, \tag{1}$$ and we observe that for any $$z \in \mathbb C$$, $$f(-z) = (-z)^{1/2} + (-(-z))^{1/2} = (-z)^{1/2} + z^{1/2} = f(z). \tag{2}$$ Therefore, if $$r$$ is a root of $$f$$, then $$-r$$ is a root of $$f$$.

Certain rules about how to manipulate functions of square roots are inherited from assumptions about the domain of such functions; e.g., $$\sqrt{ab} = \sqrt{a} \sqrt{b}$$ is true if $$a, b$$ are nonnegative real numbers. Otherwise, we encounter inconsistencies, such as the well-known $$1 = \sqrt{1} = \sqrt{(-1)(-1)} \overset{?}{=} \sqrt{-1} \sqrt{-1} = i^2 = -1. \tag{3}$$ The equality with the $$?$$ symbol above it is where the error occurs.

When we talk about square roots of complex numbers, we are really talking about a one-to-two multivalued mapping; e.g., $$(-2i)^{1/2} = \{1-i, -1+i\}.$$ This is because $$(1-i)^2 = -2i$$ and $$(-1+i)^2 = -2i$$, and because $$\mathbb C$$ is not an ordered field, unlike $$\mathbb R$$, it is not a simple matter to decide which of these roots is "canonical" in the way that we decide to use $$\sqrt{x}$$ to denote the nonnegative square root of $$x$$ when $$x$$ is a nonnegative real. Moreover, when considering cube roots, now one has in general three complex-valued solutions, all equally valid. A major fundamental aspect of complex analysis concerns itself with the choice of a single value when a mapping is multivalued.

That said, it is clear that your first step is problematic:

$$\sqrt{-x} = \sqrt{-1}\sqrt{x} = i \sqrt{x}$$ does not always hold for the same reason why the aforementioned "paradox" $$(3)$$ is invalid.

In order to proceed along the same lines of your reasoning, you must be more careful:

$$z^{1/2} + (-z)^{1/2} = z^{1/2} + (-1)^{1/2} z^{1/2} = z^{1/2} (1 + (-1)^{1/2})$$ is allowed. But here, the value of $$(-1)^{1/2}$$ must be ascertained. It is not simply $$i$$, because there are two solutions to the equation $$z^2 = -1,$$ namely $$z = \{i, -i\}.$$ Therefore, the following step is applied:

$$z^{1/2} (1 + (-1)^{1/2}) = z^{1/2} (1 \pm i). \tag{4}$$ This preserves the multivalued character of the original expression, from which we proceed as follows:

$$z^{1/2} = \frac{2}{1 \pm i},$$

hence

$$z = \left(\frac{2}{1 \pm i}\right)^2 = \{-2i, 2i\} = \pm 2i.$$

Squaring twice,

$$x +(-x) + 2 \sqrt{x}\sqrt{-x} =4 \to -x^2=4$$

$$x^2+4 = 0$$
$$x= \pm \sqrt {2} i~.$$
For the function $$\ f(x) \ = \ \sqrt{x} + \sqrt{-x} \ \ , \$$ if we are restricted to $$\ x \ \in \ \mathbb{R} \ \ , \$$ the first term is only defined for $$\ x \ \ge \ 0 \ \$$ and the second term, for $$\ x \ \le \ 0 \ \ , \$$ so the domain for $$\ f(x) \$$ is just the intersection of these intervals, $$\ x \ = \ 0 \ \ , \$$ and its only permissible value is $$\ f(0) \ = \ 0 \ \ . \$$ Consequently, to solve $$\ f(x) \ = \ r \ \ , \$$ with $$\ \ r \$$ being a non-zero real number, we are "forced onto the Argand plane" and must search for complex-valued roots of the equation.
We take a prospective root of the equation to be $$\ z \ = \ \rho·(\cos \theta \ + \ i·\sin \theta) \ = \ \rho·e^{ \ i \ · \ \theta} \ \ , \ \rho \$$ being a positive real number (the modulus of $$\ z \ ) \ \ .$$ As heropup points out, the equation $$\ z^{1/2} \ + \ (-z)^{1/2} \ = \ r \ \$$ has a symmetry in that if $$\ z \$$ is a root, $$\ (-z) \$$ is also. As complex numbers have two square-roots, we have $$z^{1/2} \ \ = \ \ \pm\sqrt{\rho} \ · \ [ \ \cos ( \theta / 2 ) \ + \ i·\sin (\theta / 2) \ ] \ \ .$$ The root $$\ (-z) \$$ lies in exactly the opposite direction from the origin that $$\ z \$$ does, so we have $$-z \ \ = \ \ \rho·[ \ \cos (\pi \ + \ \theta) \ + \ i·\sin (\pi \ + \ \theta) \ ]$$ $$\Rightarrow \ \ (-z)^{1/2} \ \ = \ \ \pm\sqrt{\rho} \ · \ [ \ \cos (\pi / 2 \ + \ \theta / 2 ) \ + \ i·\sin (\pi / 2 \ + \ \theta / 2 ) \ ] \ \ ,$$ which tells us that the square-roots of $$\ (-z) \$$ lie at equal distances from the origin and in directions perpendicular to the square-roots of $$\ z \ \ . \$$ Thus we have, $$\ z^{1/2} \ + \ (-z)^{1/2}$$ $$= \ \ \pm \sqrt{\rho} \ · \ [ \ \cos ( \theta / 2 ) \ + \ i·\sin (\theta / 2) \ ] \ + \ \pm\sqrt{\rho} \ · \ [ \ \cos (\pi / 2 \ + \ \theta / 2 ) \ + \ i·\sin (\pi / 2 \ + \ \theta / 2 ) \ ]$$ $$= \ \ \pm \sqrt{\rho} \ · \ [ \ \{ \ \cos ( \theta / 2 ) + \cos (\pi / 2 \ + \ \theta / 2 ) \ \} \ + \ i·\{ \ \sin ( \theta / 2 ) + \sin (\pi / 2 \ + \ \theta / 2 ) \ \} \ ]$$ $$= \ \ \pm \sqrt{\rho} \ · \ [ \ 2· \cos \left( \frac{\pi / 2 \ + \ \theta}{2} \right) · \cos (-\pi / 4 ) \ \ + \ i· 2· \sin \left( \frac{\pi / 2 \ + \ \theta}{2} \right) · \cos (-\pi / 4 ) \ ]$$ $$= \ \ \pm \sqrt{\rho} · \sqrt2 \ · \ [ \ \cos ( \pi / 4 \ + \ \theta / 2 ) \ + \ i· \sin ( \pi / 4 \ + \ \theta / 2 ) \ ]$$ or $$\pm \sqrt{\rho} \ · \ [ \ \cos ( \theta / 2 ) \ + \ i·\sin (\theta / 2) \ ] \ + \ \mp\sqrt{\rho} \ · \ [ \ \cos (\pi / 2 \ + \ \theta / 2 ) \ + \ i·\sin (\pi / 2 \ + \ \theta / 2 ) \ ]$$ $$= \ \ \pm \sqrt{\rho} \ · \ [ \ \{ \ \cos ( \theta / 2 ) - \cos (\pi / 2 \ + \ \theta / 2 ) \ \} \ + \ i·\{ \ \sin ( \theta / 2 ) - \sin (\pi / 2 \ + \ \theta / 2 ) \ \} \ ]$$ $$= \ \ \pm \sqrt{\rho} \ · \ [ \ -2· \sin \left( \frac{\pi / 2 \ + \ \theta}{2} \right) · \sin (-\pi / 4 ) \ \ + \ i· 2· \cos \left( \frac{\pi / 2 \ + \ \theta}{2} \right) · \sin (-\pi / 4 ) \ ] \ \ = \ \ 0$$ $$= \ \ \pm \sqrt{\rho} · \sqrt2 \ · \ [ \ \sin ( \pi / 4 \ + \ \theta / 2 ) \ - \ i· \cos ( \pi / 4 \ + \ \theta / 2 ) \ ]$$ applying here the sum-to-product identities.
We find four possible implied sums of the square-roots. When we apply our resulting expressions to $$\ z^{1/2} \ + \ (-z)^{1/2} \ = \ r \ \ , \$$ we find for $$\ r + i·0 \ > \ 0 \ \ , \$$ $$\pm \sqrt{2·\rho} \ \ = \ \ r \ \ \Rightarrow \ \ \rho \ \ = \ \ \frac{r^2}{2} \ \ ,$$ $$\cos ( \pi / 4 \ + \ \theta / 2 ) \ \ = \ \ 1 \ \ \ , \ \ \ \sin( \pi / 4 \ + \ \theta / 2 ) \ \ = \ \ 0$$ $$\Rightarrow \ \ \frac{\theta}{2} \ + \ \frac{\pi}{4} \ \ = \ \ 0 \ + \ 2k \pi \ \ \Rightarrow \ \ \frac{\theta}{2} \ \ = \ \ \frac{(8k - 1)· \pi }{4} \ \ \Rightarrow \ \ \theta \ \ = \ \ -\frac{ \pi}{2} \ + \ 4k \pi \ \ ,$$ and for $$\ r + i·0 \ < \ 0 \ \ , \$$ $$\cos ( \pi / 4 \ + \ \theta / 2 ) \ \ = \ \ -1 \ \ \ , \ \ \ \sin( \pi / 4 \ + \ \theta / 2 ) \ \ = \ \ 0$$ $$\Rightarrow \ \ \frac{\theta}{2} \ + \ \frac{\pi}{4} \ \ = \ \ \pi \ + \ 2k \pi \ \ \Rightarrow \ \ \frac{\theta}{2} \ \ = \ \ \frac{(8k + 3)· \pi }{4} \ \ \Rightarrow \ \ \theta \ \ = \ \ \frac{3 \pi}{2} \ + \ 4k \pi \ \ .$$ When $$\ r \$$ is real then, the set of roots numbers just two, $$\ z \ = \ -\frac{r^2}{2}·i \ \$$ and, due to the symmetry of $$\ f(x) \ \ , \ (-z) \ = \ +\frac{r^2}{2}·i \ \ .$$ For the specific equation of this problem, the roots are $$\ \pm \ \frac{2^2}{2}·i \ = \ \pm \ 2i \ \ . \$$ If we write the square-roots of $$\ z \$$ as $$\ u_{\pm} \$$ and those of $$\ (-z) \$$ as $$\ v_{\pm} \ \ , \$$ the graph below shows that positive and negative values of $$\ r \$$ are obtained by choosing different sums of square-roots.