Polar coordinates in a 2d complex number space

Question

A quick disclaimer that I'm working on all of this for recreational reasons, so I might be missing a lot of the precise language. I haven't taken any complex analysis, yet. (really excited for it, however)

Also, real quickly, I'm not asking about how to represent a complex number in polar form i.e. representing $a+bi$ as $re^{i\theta}$. (I just don't want to get marked as a duplicate right off the bat and I couldn't find this question among all those.)

Consider $\mathbb C^2$. Does it make sense to talk about transforming a point, $(x,y)$ into a polar form $(\theta,r)$? To begin, I decided to consider the analog of the unit circle in this space (the set of all points one unit from the origin). For this I went with the usual distance function, $r=\sqrt{x^2+y^2}$.

Ok, lets think about the point on this unit "circle" that has a y-coordinate of 2. Applying the distance formula above, this means that $x=\pm\,i\sqrt{3}$. I'll take the positive value. Then by analogy to the familiar unit circle, assume that $y=\sin{\theta}=2$, which gives $\theta = \frac{\pi}{2}-i\, \mathrm{arccosh}\,2$. I've been able to convince myself that the real part of this angle makes sense as $\frac{\pi}{2}$ in this context:

Most fundamentally, sin is the ratio of the opposite side and hypotenuse of a right triangle. If we take the modulus of the side lengths, we get, in this case, $2, \sqrt{3}$, and $1$, but the right angle, is positioned at the angle of interest. (I hope that makes sense it's hard to explain without a picture.)

It's a little harder for me to get my hands on the imaginary part of the angle. I assume that it's related to some hyperbolic stuff that I just haven't learned yet. I know that there's a way to give a meaning to $\mathrm{arccosh}\,2$ as some sort of hyperbolic angle, but for me the intuition isn't there.

My question then, is: What does the complex part of the angle represent intuitively?

Also does any of what I've done really follow/make sense? I'm particularly concerned about the way I defined my metric--it is defined such that a point with coordinates who's moduli are arbitrarily large can still be one unit from the origin. Then again so it is with the so-called unit hyperbola if I understand properly.

And finally, how might we visualize this unit circle (hyper-sphere)?

EDIT: Thank you very much for the comments and answers. I thought I would change up my question so that I can either get answers that are closer to my knowledge base or that answer my question as I clarify it here if the current answers do misunderstand me.

Essentially what I am considering is a "right" triangle which can have complex side lengths. In particular, I was considering the triangle formed from sides of $2$,$1$, and $\sqrt{3}\,i$ side lengths. It seems natural to say that the angle between the shortest two sides (in absolute value) is $\theta = \frac{\pi}{2}-i\, \mathrm{arccosh}\,2$, and then I was asking, what does the imaginary part of that angle represent. I'm not asking about plugging in a complex value to $e^{i \theta}$, which from what I understood of the answers, was what I was taken to be asking.

Answer 1

"Complex length" is something of an oxymoron: a length in geometry is real, mostly because we want lengths to be absolutely comparable (to take values in an ordered field), and we want a limit of lengths to be a length (completeness).

Consequently, trying to extend the circular functions to a complex setting as ratios of sides of right triangles isn't obviously meaningful. (You can think of sides of a real plane triangle as complex numbers, but then the (non-real!) complex ratios are not the usual sine or cosine; matters are complicated further if you move into the realm of complex vectors.)

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If instead you mean to complexify polar coordinates, i.e., to write $$ x = (r + is) \cos(\theta + i\phi),\qquad y = (r + is) \sin(\theta + i\phi) $$ with $r$, $s$, $\theta$, and $\phi$ real, and to view $\phi$ as the imaginary part of the angle, then the sum-of-angles formulas for sine and cosine, and the identities $$ \cos(i\phi) = \cosh\phi,\qquad \sin(i\phi) = i\sinh\phi $$ give \begin{align*} x &= (r + is) \cos(\theta + i\phi) \\ &= (r + is)(\cos\theta \cosh\phi - i\sin\theta \sinh\phi) \\ &= (r\cos\theta \cosh\phi + s\sin\theta \sinh\phi) + i(s\cos\theta \cosh\phi - r\sin\theta \sinh\phi) \end{align*} and \begin{align*} y &= (r + is) \sin(\theta + i\phi) \\ &= (r + is)(\sin\theta \cosh\phi + i\cos\theta \sinh\phi) \\ &= (r\sin\theta \cosh\phi - s\cos\theta \sinh\phi) + i(s\sin\theta \cosh\phi + r\cos\theta \sinh\phi). \end{align*}

To focus on the "imaginary part of the angle" $\phi$, you might write $$ \left[\begin{array}{@{}c@{}} x \\ y \\ \end{array}\right] = (r + is)\left\{ \cosh\phi \left[\begin{array}{@{}c@{}} \cos\theta \\ \sin\theta \\ \end{array}\right] + i\sinh\phi \left[\begin{array}{@{}r@{}} -\sin\theta \\ \cos\theta \\ \end{array}\right]\right\} $$ and think about what happens when $\theta$ is fixed and $\phi$ allowed to vary. I don't see any nice intuitive interpretation offhand, however. (The two real unit column vectors are orthogonal, and orthogonality is maintained when the second is multiplied by the imaginary unit. The expression is braces therefore parametrizes a hyperbola in the plane the two vectors span.)

In case it's of interest, my answer to Parametric equation - of a hyperbola contains a couple of diagrams giving a geometric analogy between the circular and hyperbolic functions:

Answer 2

Suppose that $\theta = \omega + i\beta$, where $\omega$ is the angle's real part and $i\beta$, its imaginary part.

Let us build up an answer by stages.

1. Let $z_1 = e^{i\omega} = \cos\omega + i\sin\omega$. This much is intuitively clear to you, is it not? The $\omega$ is just an angle in the Argand plane.
2. Let $z_2 = e^{-\beta}$. In this case, the $\beta$ is just an argument of decay.
3. Let $z_3 = e^{-\beta}e^{i\omega}$. This is just a combination of the other two. You have the angle, $\omega$, and the argument of decay, $\beta$.
4. Let $z_4 = e^{i\theta} = e^{i(\omega+i\beta)}$. What does this mean? Answer, it means the same as $z_3$. Indeed, $z_4=z_3$.

Therefore, an angle's imaginary part may be said to represent an argument of decay that is to be associated with the real-valued angle.

Regarding the hypersphere, though it is attractive to suppose that $\mathbb C^2$ were four-dimensional, that view does not work out well in practice. Rather, it's two-dimensional with complex elements. Where you really see this is in $\mathbb C^3$, in which, say, the spatial Fourier transform of a magnetic field is represented. Maybe you haven't studied Fourier, yet, but when you do, you'll see that it just does not make much sense to think of the three-dimensional complex field as though it were six-dimensional.

Answer 3

I doesn't work like that. We never encounter imaginary angles in complex variables. Okay, suppose you concoct an imaginary $\theta=\alpha+i\beta$, then your complex number would be, say

$$z=e^{i\theta}=e^{i\alpha-\beta}=e^{-\beta}e^{i\alpha}$$

This is, in fact, a logarithmic spiral with a flair coefficient of $-\beta$, i.e., it's shrinking.

Okay, back to the main point. Say we have a unit circle at the origin, given by $z=e^{i\theta}$ and we want to move it to $a+ib$. What becomes of $z$? Well, we have

$$z=e^{i\theta}+a+ib=(a+\cos\theta)+i(b+\sin\theta)=re^{i\varphi}\\ r=|z|=\sqrt{(a+\cos\theta)^2+(b+\sin\theta)^2}\\ \varphi=\tan^{-1}\left(\frac{\Im\{z\}}{\Re\{z\}} \right) $$

There's no imaginary part to the angle. To state it another way, a point in the complex plane has a distance $r$ from the origin and an angle $\theta$ anticlockwise from the $x$-axis. An imaginary angle would simply tell you that the point is somewhere else.