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We know that any complex number in its cartesian representation can be expressed as:

  • z = x + i y

But if we express it in its complex exponential form we have:

  • z = r exp(i teta)

In this last complex exponential form, we are actually expressing not just the complex number z = x + i y, but an infinite quantity of complex numbers, all the ones that share the same radius r and that have same angle but different by an 2 k pi, being k integer. Now, you could say: all the ones in exponential form correspond to the same point in the complex plane so what's the matter? Ok, but mathematically speaking they are conceptually separate points which overlap each others, since they correspond to different polar cohordinates!

e.g. z = i

cartesian: (0, i) complex exponential: (1, pi/2), (1, pi/2 + 2pi), (1, pi/2-2pi), ...

Therefore question is: are cartesian and exponential forms actually equivalent representations of the same complex number? I ask this because it doesn't seem to exist a one to one correspondence between any couple (x, y) and only one couple (r, teta) and viceversa in the complex plane...

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    $\begingroup$ If my sister calls me "brother" and my mother calls me "son", am I conceptually two different people? $\endgroup$
    – JonathanZ
    Jul 17, 2022 at 22:45
  • $\begingroup$ @JonathanZsupportsMonicaC no, but if your sister calls you "brother" and your mom calls you "son+2kpi" i really don't know who's she referring to! haha $\endgroup$
    – Giack_89
    Jul 17, 2022 at 22:53
  • $\begingroup$ Is $\frac{2}{2}$ equivalent to $\frac{3}{3}$? $\endgroup$ Jul 18, 2022 at 0:19
  • $\begingroup$ @user2661923 yes it is... but 2/2 and 3/3 are the very same number, i.e. 1. Considering instead for example (r, pi/2) and (r, pi/2+2pi), they are not the same complex number because they are two points with different cohordinates in the polar plane. $\endgroup$
    – Giack_89
    Jul 18, 2022 at 9:12
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    $\begingroup$ @Giack_89 Wrong. They have different polar coordinates, but are still the exact same number. This can be established by computing the real and imaginary components of the two numbers. $\endgroup$ Jul 18, 2022 at 9:13

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I don't understand what you mean by "equivalent" but you are correct that there is not a bijective correspondence between cartesian coordinates $(x, y)$ and polar coordinates $(r, \theta)$. In addition to the $2 \pi i k$ ambiguity you describe there is a second much worse ambiguity when $r = 0$, where the origin $0 + 0 i$ can have polar coordinates $0 e^{i \theta}$ for any $\theta$.

After deleting the origin only the $2 \pi i k$ ambiguity remains. There are a couple of ways to talk about this. One is to restrict $\theta$ to a half-open interval, say $[0, 2\pi)$ or $[-\pi, \pi)$. Then it's unique but this is conceptually unsatisfying because then the map from cartesian to polar coordinates fails to be continuous once we hit the endpoints. Another is to say that $\theta$ is not really a real number but actually an element of the quotient space $\mathbb{R} / 2 \pi \mathbb{Z}$ where we identify two real numbers if they differ by an integer multiple of $2 \pi$; then the exponential map describes an isomorphism between this quotient space and the circle $S^1$.

A third (closely related to but not quite the same as the second) is to talk about covering spaces and to say that the polar coordinate map $(r, \theta) \mapsto re^{i \theta}$ really defines a covering map from $\mathbb{R}_{+} \times \mathbb{R}$ to $\mathbb{C}^{\times} \cong \mathbb{C} \setminus \{ 0 \}$; the existence of this covering map abstractly reflects the fact that the fundamental group of the punctured complex plane is nonzero. This leads to winding numbers and Riemann surfaces and lots of other fun stuff.

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    $\begingroup$ By "not equivalent" i meant that there is not a bijective correspondence between the two representations, as you also agreed. First because, as you said, for r = 0, teta can be any, and then because of the 2*k*pi problem. You also posted a lot of references i've never heard about (e.g. covering spaces, quotient spaces, Rieman surfaces, etc), thank you, i'm going to give a read hoping is not too hard for my engineering skills :). Anyway i'm glad at least that actually is it an actual problem and it's not that i just went mad! $\endgroup$
    – Giack_89
    Jul 17, 2022 at 23:21

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