number of solutions of x if x^2=a+ib?

Question

We know that if $x^2=a,a\in\mathbb{R}$ $$|x|=\sqrt a \quad \implies \quad x=\pm\sqrt a$$ Hence we have 2 possible values for $x$. Now if $x^2=a+ib$ $$|x|=\sqrt{a+ib} \quad\implies\quad x=\pm\sqrt{a+ib}$$ Then how many values of $x$ are possible?

Answer 1

So we are looking for solutions of $x^2-c=0$ over $\Bbb C$. There are 2 cases:

1. If the discriminant $D=4c$ of the quadratic polynomial vanishes, then there is only one solution with multplicity 2: Zero is a root of order 2 of $x^2=0$.
2. If the discriminant is non-zero (i.e. if $c\neq 0$), there are two distinct solutions which are $\sqrt c$ and $-\sqrt c$ where $\sqrt{\ }$ denotes the principal branch of the complex square root.

If $c$ is given in cartesian coordinates like $c=x+iy$ with $x,y\in\Bbb R$, see here for a representation of the value of the (principal branch of the) square root at $c$.

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$$|x|=\sqrt{a+ib} \quad\implies\quad x=\pm\sqrt{a+ib}$$

This is no more all solutions over the complex numbers: if $|x| = r$, then $x$ may lay anywhere on the circle of radius $r$ centered at the origin: Any $x$ of the form $$x=r\cdot e^{iw} = r(\cos w + i\sin w)\ ,\ w\in\Bbb R$$ satisfies $|x|=r$.

Answer 2

Given that we are working in the field of complex numbers $\mathbb{C},$ we have available to us the complex conjugate function $^*.$ This means that $z^*$ is well-defined. As such, we can define the function $\mathfrak{Re}:\mathbb{C}\to\mathbb{R}$ such that $$\mathfrak{Re}(z)=\frac{z+z^*}2$$ and $\mathfrak{Im}:\mathbb{C}\to\mathbb{R}$ such that $$\mathfrak{Im}(z)=\frac{z-z^*}{2i}.$$ As such, one has that $$z=\mathfrak{Re}(z)+\mathfrak{Im}(z)i$$ for all $z\in\mathbb{C}.$ Also, one has the absolute value function $||:\mathbb{C}\to[0,\infty)$ with $|z|=\sqrt{zz^*}.$ Finally, we have the $\mathrm{atan2}:(\mathbb{R}-\{0\})^2\to(-\pi,\pi]$ defined as it here in this Wikipedia article https://en.wikipedia.org/wiki/Atan2. We can define the function $\theta:\mathbb{C}-\{0\}\to(-\pi,\pi]$ as $$\theta(z)=\mathrm{atan2}[\mathfrak{Im}(z),\mathfrak{Re}(z)],$$ and with this, we can say $$z=|z|e^{\theta(z)i}.$$ Why do I do this? Because now, for the equation $z^2=a$ for $a\in\mathbb{C},$ we have $$|z|^2e^{2\theta(z)i}=|a|e^{\theta(a)i}.$$ Thus $$|z|^2=|a|$$ and $$e^{2\theta(z)i}=e^{\theta(a)i}.$$ The first equation is easy to solve for $|z|,$ leaving us with $|z|=\sqrt{|a|}.$ The second equation leaves us with $$\theta(z)=\frac{\theta(a)+2m\pi}2=\frac{\theta(a)}2+m\pi$$ where $m\in\mathbb{Z}.$ Therefore, $$z=\sqrt{|a|}e^{\frac{\theta(a)}2i+m\pi{i}}=(-1)^m\sqrt{|a|}e^{\frac{\theta(a)}2i}.$$ As a matter of notational extension, for $a\in\mathbb{C},$ we can define $$\sqrt{a}=\sqrt{|a|}e^{\frac{\theta(a)}2i},$$ so that $z=(-1)^m\sqrt{a},$ hence $z=-\sqrt{a}$ or $z=\sqrt{a}.$ Notice, though, that $|z|=a$ does not imply $z=\pm{a},$ it implies $z=ae^{ti}$ for $t\in(-\pi,\pi].$ Also, $\sqrt{z^2}=|z|$ is false. In reality, $$\sqrt{z^2}=\begin{cases}z&\theta(z)\in\left(-\frac{\pi}2,\frac{\pi}2\right]\\-z&\mathrm{otherwise}\end{cases}.$$ One also can generalize this for higher degrees. If one consider $z^n=a,$ then one can say $$\sqrt[n]{a}=\sqrt[n]{|a|}e^{\frac{\theta(a)}{n}i},$$ so that $$z=e^{\frac{2m\pi{i}}n}\sqrt[n]{a}$$ for $m\in\mathbb{Z}$ gives the full family of solutions. One can even choose to define $$\omega_n=e^{\frac{2\pi{i}}{n}},$$ as is traditional to do with roots of unity, so that $z^n=a$ is equivalent to $z=\omega_n^m\sqrt[n]{a}$ for $m\in\mathbb{Z}.$ This is the correct way to approach the problem.