Solving complex equation, multiple solutions

Question

I was tying to solve (1) $\sqrt{x} + \sqrt{-x} = 1$

$\sqrt(x)(1+i) = 1 \rightarrow \sqrt(x) = \frac{1}{1+i} \rightarrow x=-\frac{1}{2}i$

One can see that by symmetry $x=\frac{1}{2}i$ is also a solution.

Question1, Why doesnt solving (1) directly give two solutions?

Now check by filling in x in (1)

$x=\frac{1}{2}i$ can be written as $\frac{1}{2}e^{i\frac{1}{2}\pi}$

$x=-\frac{1}{2}i$ can be written as $\frac{1}{2}e^{i\frac{3}{2}\pi}$ or as $\frac{1}{2}e^{i\frac{-1}{2}\pi}$

$\sqrt{\frac{1}{2}e^{i\frac{1}{2}\pi}} + \sqrt{\frac{1}{2}e^{i\frac{-1}{2}\pi}} = \sqrt\frac{1}{2}.(e^{i\frac{1}{4}\pi}+e^{i\frac{-1}{4}\pi}) = \sqrt\frac{1}{2}.( (\sqrt\frac{1}{2}+i\sqrt\frac{1}{2})+ (\sqrt\frac{1}{2}-i\sqrt\frac{1}{2})) = 1$ OK

But using $\frac{1}{2}e^{i\frac{3}{2}\pi}$

$\sqrt{\frac{1}{2}e^{i\frac{1}{2}\pi}} + \sqrt{\frac{1}{2}e^{i\frac{3}{2}\pi}} = \sqrt\frac{1}{2}.(e^{i\frac{1}{4}\pi}+e^{i\frac{3}{4}\pi}) = \sqrt\frac{1}{2}.( (\sqrt\frac{1}{2} +i\sqrt\frac{1}{2})+ (-\sqrt\frac{1}{2}+i\sqrt\frac{1}{2})) = i$ NOT OK

Question2, What am i doing wrong here, this probably has someting to do with "primitive roots". But I dont see it.

Answer 1

Both (1) and (2) have the same explanation: The definition of $\sqrt{a}$ for complex $a$ depends on a choice of branch, and your computation assumes a particular choice. For another choice, we would have $\sqrt{-1} = -i$. Analyzing all the possibilities gives two solutions, $x = \pm \frac{i}2$.

Answer 2

Let $\pm z_0$ be the two solutions of $z^2=x$

Then the two solutions of $z^2=-x$ will be $\pm iz_0$

So in the end you have $4$ equations to solve $(\pm 1+\pm i)z_0=1\iff z_0=\pm\frac 12+\pm\frac i2$ and not just the single one you indicated i.e $z_0(1+i)=1$

These $4$ solutions for $z_0$ gives only $2$ solutions for $x$ once squared back, $x=(z_0)^2=\pm\frac i2$

Now the issue you have when reporting the values in the equation is that in this case $\pm\frac 12+\pm\frac i2$ gives 4 numbers all in a different region of the complex plane (in 4 "quadrants" if you prefer). But there is no $\sqrt{\cdot}\ $ that can satisfy all these regions at once with the same branch cut.