# Visualizing and understanding the roots of $f(z) = z^2 - e^{i\varphi}$

[I've added another animated picture below, showing how the actions of the groups $Q_\alpha$, $R_\beta$ coincide every now and then.]

The roots of $f(z) = z^2 - e^{i\varphi}$ are simply $\pm e^{i\varphi/2}$. They can be visualized as the intersection of two curves in the complex plane (see my other question here):

$\color{red}{C^\varphi_r = \{ (x,y)\ |\ x^2 - y^2 - \cos\varphi = 0\}}$

$\color{green}{C^\varphi_i = \{ (x,y)\ |\ 2xy - \sin\varphi = 0\}}$

So we have

$\color{red}{y_r^\varphi(x) = \pm \sqrt{x^2 - \cos\varphi}}$

$\color{green}{y_r^\varphi(x) = \sin\varphi/2x}$ (at least for $\varphi \neq 0$)

The black dot represents $\varphi$, the blue dots represent the two roots of $f(z) = z^2 - e^{i\varphi}$:

There are four interesting special cases in which straight lines occur:

• $\varphi = 0$, i.e. $z^2 = 1$ with
$\color{red}{y_r^0(x) = \pm \sqrt{x^2 - 1}}$
$\color{green}{C^0_i = \{ (x,y)\ |\ x = 0 \vee y = 0\}}$

• $\varphi = \pi/2$, i.e. $z^2 = i$ with
$\color{red}{y_r^{\pi/2}(x) = \pm x}$
$\color{green}{y_i^{\pi/2}(x) = 1/2x}$

• $\varphi = \pi$, i.e. $z^2 = -1$ with
$\color{red}{y_r^{\pi}(x) = \pm \sqrt{x^2 + 1}}$
$\color{green}{C^\pi_i = \{ (x,y)\ |\ x = 0 \vee y = 0\}}$

• $\varphi = 3\pi/2$, i.e. $z^2 = -i$ with
$\color{red}{y_r^{\pi/2}(x) = \pm x}$
$\color{green}{y_i^{\pi/2}(x) = -1/2x}$

What may come as a surprise is the fact that the four pictures seem to be related by rotations by $k\pi/4$, at least when ignoring the colors. As it turns out, this impression is correct, i.e. the set $\color{red}{C^0_r}$ rotated by $\alpha = (2k+1)\pi/4$ coincides with $\color{green}{C^\beta_i}$ with $\beta = (2k+1)\pi/2$.

My questions are:

• Does an expericenced mathematician see at a glance that the functions $f(x) = \sqrt{x^2 + 1}$ and $g(x) = 1/2x$ have the very same shape?

• How does he prove it? (I did not manage.)

• Are there general symmetry arguments that tell us, that $\color{red}{C^0_r}$ must have the same shape as $\color{green}{C^{\pi/2}_i}$ (so it has not to be proved by solving complex equations).

The observation can be summed up this way: Let $C^\varphi = C^\varphi_r \cup C^\varphi_i$. For the two groups of transformations $Q_\alpha$, $R_\beta$ with

• $Q_\alpha (C^\varphi) = C^{\varphi + \alpha}$

• $R_\beta (C^\varphi) = \text{ rotation of } C^{\varphi} \text{ by } \beta$

$Q_\alpha (C^\varphi) = R_\beta(C^\varphi)$ holds iff $\alpha = k\pi/2$ and $\beta = \alpha/2$ (is this correct?) which somehow would reflect the fact that the roots of $f(z) = z^2 - e^{i\varphi}$ are given by $\pm e^{i\varphi/2}$:

What can be learned from this observation?

Side question (see my related question):

Let $\color{red}{S^\varphi_r}$ be the graph of the real part of $f^\varphi(z) = z^2 - e^{i\varphi}$, i.e. $\color{red}{S^\varphi_r} = \{(x,y,z)\ | \ (x,y,x^2 - y^2 - \cos\varphi), x,y \in \mathbb{R}\}$

Let $\color{green}{S^\varphi_i}$ be the graph of the imaginary part of $f^\varphi(z) = z^2 - e^{i\varphi}$, i.e. $\color{green}{S^\varphi_i} = \{(x,y,z)\ | \ (x,y,2xy - \sin\varphi), x,y \in \mathbb{R}\}$

$\color{red}{S^\varphi_r}$ and $\color{green}{S^\varphi_i}$ are 2-dimensional surfaces in $\mathbb{R}^3$. Their intersections with the $(x,y)$ plane are $\color{red}{C^\varphi_r}$ and $\color{green}{C^\varphi_i}$.

Consider $\color{blue}{C^\varphi} = \color{red}{S^\varphi_r} \cap \color{green}{S^\varphi_i}$. How does $\color{blue}{C^\varphi}$ look like? (I have no idea. Maybe some circles or other conic sections?)

[You can play around with $f(z) = z^2 - e^{i\varphi}$ here. I hope it works.]

• Are you familiar with linear algebra? – Servaes Aug 23 '18 at 17:26
• How does linear algebra answer my questions? (Please have a look at my questions.) – Hans-Peter Stricker Aug 23 '18 at 17:55
• I admit I should know how to rotate a point set and see if it equals another one. But as I said: I did not manage. – Hans-Peter Stricker Aug 23 '18 at 17:58
• I'll illustrate how to, using some basic linear algebra. – Servaes Aug 23 '18 at 17:58
• Great, I'm looking forward to it! (Having said this: Yes, I'm familar with linear algebra!) – Hans-Peter Stricker Aug 23 '18 at 17:59

Your question raises a few closely related questions. I've tried to adress all of them, though not all in the same amount of detail. Let me know if you are missing something here.

You've defined two families of curves $\color{red}{C_r^{\varphi}}$ and $\color{green}{C_i^{\varphi}}$, parametrized by $\varphi\in[0,2\pi)$. I'll omit the colours for my own convenience, and in stead define $$A(\varphi):=\color{red}{C_r^{\varphi}}\qquad\text{ and }\qquad B(\varphi):=\color{green}{C_i^{\varphi}}.$$ Now for any given value of $\varphi$, the curves $A(\varphi)$ and $B(\varphi)$ are the graphs (in the $(x,y)$-plane) of the functions $$f_{\varphi}(x,y):=x^2-y^2-\cos\varphi\qquad\text{ and }\qquad g_{\varphi}(x,y):=2xy-\sin\varphi,$$ respectively. Consider a rotation of the plane around the origin over an angle $\theta$, let's denote it by $R_{\theta}$ as in your question. It sends a point $(x,y)$ to $(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$. Using matrix-vector notation this is concisely written as $$R_{\theta}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\cos\theta-y\sin\theta\\x\sin\theta+y\cos\theta\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}.$$ In particular, for $\theta=\tfrac{\pi}{4}$ we see that a point $(x,y)$ maps to $$R_{\frac{\pi}{4}}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\\\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}x-y\\x+y\end{pmatrix}.$$ If $(x,y)\in B(\varphi)$ then $g_{\varphi}(x,y)=0$. But then \begin{eqnarray*} g_{\varphi}(R_{\theta}(x,y)) &=&g_{\varphi}\left(\frac{x-y}{\sqrt{2}},\frac{x+y}{\sqrt{2}}\right) =2\frac{x-y}{\sqrt{2}}\frac{x+y}{\sqrt{2}}-\sin\varphi\\ &=&(x-y)(x+y)-\sin\varphi=x^2-y^2-\cos(\varphi-\tfrac{\pi}{2})\\ &=&f_{\varphi+\tfrac{\pi}{2}}(x,y), \end{eqnarray*} so we see that the curve $A(\varphi-\tfrac{\pi}{2})$ is simply the curve $B(\varphi)$ rotated over an angle of $\theta=\tfrac{\pi}{4}$. Repeating the same argument also shows that${}^1$ the curve $B(\varphi-\tfrac{\pi}{2})$ is simply the curve $A(\varphi)$ rotated over an angle of $\theta=\frac{\pi}{4}$: \begin{eqnarray*} f_{\varphi}(R_{\theta}(x,y)) &=&f_{\varphi}\left(\frac{x-y}{\sqrt{2}},\frac{x+y}{\sqrt{2}}\right) =\left(\frac{x-y}{\sqrt{2}}\right)^2-\left(\frac{x+y}{\sqrt{2}}\right)^2-\cos\varphi\\ &=&\frac{x^2-2xy+y^2}{2}-\frac{x^2+2xy+y^2}{2}-\cos\varphi=-2xy+\sin(\varphi-\tfrac{\pi}{2})\\ &=&-g_{\varphi+\tfrac{\pi}{2}}(x,y), \end{eqnarray*} and of course $-g_{\varphi+\tfrac{\pi}{2}}(x,y)=0$ if and only if $g_{\varphi+\tfrac{\pi}{2}}(x,y)=0$. This shows that rotating a picture of the curves $A(\varphi)$ and $B(\varphi)$ one eights of a turn simply gives you the picture of $A(\varphi-\tfrac{\pi}{2})$ and $B(\varphi-\tfrac{\pi}{2})$. In other words, going from $\varphi$ to $\varphi-\tfrac{\pi}{2}$ corresponds to turning the picture over an angle of $\theta=\tfrac{\pi}{4}$ and switching the colours. Forgetting about colours, we can say in terms of your functions $Q_{\alpha}$ and $R_{\beta}$ that $$Q_{k\tfrac{\pi}{2}}(C^{\varphi}) =C^{\varphi+k\tfrac{\pi}{2}} =A(\varphi+k\tfrac{\pi}{2})\cup B(\varphi+k\tfrac{\pi}{2}) =R_{k\tfrac{\pi}{4}}(C^{\varphi}).$$

About the functions $f(x)=\sqrt{x^2+1}$ and $g(x)=\frac{1}{2x}$, it becomes clear that their graphs are closely related once you write down the implicit equations that they are branches of; they come from the implicit functions $$(x-y)(x+y)=y^2-x^2=1\qquad\text{ and }\qquad 2xy=1.$$ In both cases a product must equal $1$, and a simply linear transformation turns one into the other; the mapping $(x,y)\ \longmapsto\ \tfrac{1}{\sqrt{2}}(x-y,x+y)$ is the 'nicest' one of many linear transformations that work. And with a little experience in linear algebra of the (real) plane, one quickly recognizes that this is a rotation over an angle of $\theta=\tfrac{\pi}{4}$.

1. This also follows from noting that both $A(\varphi)$ and $B(\varphi)$ are symmetric about the origin, i.e. that for all $x$, $y$ and $\varphi$ we have $$f_{\varphi}(x,y)=f_{\varphi}(-x-y) \qquad\text{ and }\qquad g_{\varphi}(x,y)=g_{\varphi}(-x-y).$$

Addendum: Concerning the side question about the real and imaginary parts of $f^{\varphi}(z)$ being equal: A point $(x,y,z)\in\Bbb{R}^3$ is on both surfaces if and only if $$z=x^2-y^2-\cos\varphi=2xy-\sin\varphi.$$ In particular, the $x$ and $y$ coordinates satisfy $$x^2-2xy-y^2=\cos\varphi-\sin\varphi,$$ where the right hand side is a constant between $-\sqrt{2}$ and $\sqrt{2}$. This defines a hyperbola, unless $\cos\varphi-\sin\varphi=0$, i.e. unless $\varphi=\tfrac{\pi}{4}+k\pi$ for some integer $k$. So the projection of $\color{blue}{C^{\varphi}}$ onto the $(x,y)$-plane is a hyperbola, or a union of two lines.

Note that it is generally not a conic section; it is the intersection of two quadratic surfaces, whereas a conic section is the intersection of a quadratic surface and a linear surface. The intersection is a space curve that it not contained in a plane, but the equations a bove allow for a parametric representation, for example $$x\ \longmapsto\ (x,-x\pm \sqrt{2x^2+\cos\varphi-\sin\varphi},-2x^2\pm x\sqrt{2x^2+\cos\varphi-\sin\varphi}-\sin\varphi),$$ where the $\pm$-signs correspond to the two branches of the hyperbola.

• Thanks a lot for this concise answer, wonderful. Maybe you want to have a look at another animated picture I created and inserted in my question. – Hans-Peter Stricker Aug 24 '18 at 7:30