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(Notation: define $\mathop{\mathrm{cis}}\theta = e^{i\theta} = \cos \theta + i \sin \theta)$.

I want to show that $$z^3=8i \iff z \in \{\sqrt{3}+i,-\sqrt{3}+i,-2i\}$$ using polar coordinates.

I can do it, but I can't do it rigorously.

Or at least, the logic isn't transparent to me.

Here's the solution: assume $z=r \mathop{\mathrm{cis}}\theta.$ Then, with a bit of work, we can show:

$$z^3 = 8i \iff (r=2) \mathrel{\&} \left(\theta \in \frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right)$$

It follows that

$$z^3 = 8i \iff z \in 2 \mathop{\mathrm{cis}}\left(\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right),$$

from which the desired result follows.

The trouble is, I don't really know why the italicized it follows that is justified.

Question. What principle(s) of logic are at work here, justifying the "it follows that"?

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    $\begingroup$ The only hidden assumption that I see is that every complex number can be written as $r\mathop{cis}\theta$. $\endgroup$ – Carsten S Nov 19 '16 at 14:09
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    $\begingroup$ @goblin Well, if $r=2$ and $\theta\in {\pi\over 6}+{2\pi\over 3}\mathbb{Z}$, then - since $z=r\cdot cis(\theta)$ - this means $z\in\{2\cdot cis(\theta): \theta\in \{{\pi\over 6}+{2\pi\over 3}\mathbb{Z}\}\}$, and this is exactly what the notation "$z\in 2\cdot cis({\pi\over 6}+{2\pi\over 3}\mathbb{Z})"$ means. Or am I missing something? $\endgroup$ – Noah Schweber Nov 20 '16 at 2:18
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    $\begingroup$ @goblin: the tag wiki for "logic" specifically explain that the tag is not intended for general applications of logical reasoning in other areas of math - it is specifically for the subfield of "mathematical logic". math.stackexchange.com/questions/tagged/logic This does not mean that the question is not about logical reasoning, only that it is not about the disciplines that are known as "mathematical logic". I have added a proof-writing tag, as the question seems to be about how to formulate a proof. $\endgroup$ – Carl Mummert Nov 20 '16 at 4:03
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    $\begingroup$ @goblin: as dxiv (and other answers)indicate in answers, there doesn't appear to be any inference rule of first order logic being used, just two different notations for the same set. The set of $z$ such that $r(z) = 2$ and $\theta(z) \in \frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}$ is the same as the set $2\text{cis}( \frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z} )$. To prove this in first-order logic, we would need to begin by specifying which formal deductive system we wanted to use, and the set of formal axioms that apply (e.g are we working in set theory, or in the theory of the field $\mathbb{C}$. $\endgroup$ – Carl Mummert Nov 20 '16 at 15:58
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    $\begingroup$ The "logic" is no more or less than the same logic of: Okay, class. We are going to a farm. You are going to see an animal. It is going to be a certain color. You are going to take out your pens and write: "I saw a/an < color > < animal >". Okay, Here's the animal: a yellow chicken runs by. Okay, what color was it? yellow And what animal was it? a chicken. So write it down: "I saw a yellow chicken"... a complex number is in the form z= r cis theta; r = 2, theta in pi/6 + 2pi/3 Z. So the complex number is z in 2 pi/6 + 2pi/3 Z. There's no "logic" It's just notation. $\endgroup$ – fleablood Nov 20 '16 at 17:01
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I'm not sure if this meets your requirement for being about 'logic' (although at least it is not about 'proof writing'), but one can give a rule, of which your 'it follows that' is one application, as follows.

If $f: X \to U$ is a function, then for any subset $A \subseteq X$ write $$ f[A] = \{ f(x) : x \in A \}. $$ Then for all $x \in X$, $$ f[\{x\}] = \{f(x)\}. $$

Also if $p: U \to W$ is a function, then for all $A \subseteq X$, $$ (p \circ f)[A] = p[f[A]]. $$

If $f: X \to U$ and $g: Y \to V$ are functions, define the function $$ f \times g: X \times Y \to U \times V,\ (x, y) \mapsto (f(x), g(y)). $$

Then for all $A \subseteq X$ and $B \subseteq Y$, $$ (f \times g)[A \times B] = \{ (f(x), g(y)) : x \in A \text{ and } y \in B \} = f[A] \times g[B]. $$ Therefore if $p: U \times V \to W$ is a function, then $$ \tag{*}\label{eq:rule} (p \circ (f \times g))[A \times B] = p[(f \times g)[A \times B]] = p[f[A] \times g[B]]. $$

In particular, for all $x \in X$ and $B \subseteq Y$, $$ (f \times g)[\{x\} \times B] = \{f(x)\} \times g[B] = \{ (f(x), g(y)) : y \in B \}, $$ and $$ (p \circ (f \times g))[\{x\} \times B] = p[\{f(x)\} \times g[B]] = \{ p(f(x), g(y)) : y \in B\}. $$ So my suggested 'rule', a special case of the more general 'rule' \eqref{eq:rule}, is $$ \boxed { (p \circ (f \times g))[\{x\} \times B] = \{ p(f(x), g(y)) : y \in B\}. } $$

For your application of this rule, take:

  • $X = U = \mathbb{R}_{>0}$, and $f: X \to U$ the identity function, $r \mapsto r$;

  • $Y = \mathbb{R}$, $V = \mathbb{T}$ (the unit circle in the complex plane), and $g: Y \to V$, $\theta \mapsto \operatorname{cis}\theta$;

  • $W = \mathbb{C}^*$ (the punctured complex plane, $\mathbb{C} \smallsetminus \{0\}$), and $p: U \times V \to W$, $(r, u) \mapsto ru$ (ordinary complex multiplication);

  • $x = 2$;

  • $B = \frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}$.

Polar coordinates $(r, \theta)$ represent the non-zero complex number: $$ r\operatorname{cis}\theta = f(r)g(\theta) = p(f(r), g(\theta)) = p((f \times g)(r, \theta)), $$ and the 'rule' in this instance states: $$ (p \circ (f \times g))\left[\{2\} \times \left(\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right)\right] = \left\{ 2\operatorname{cis}\theta : \theta \in \frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z} \right\}, $$ which seems (doesn't it?) to be the inference you are making.

tumbleweed

(I'll get me coat ...)

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Um, I'm a little confused by your question and your confusion.

Isn't the statement:

  1. $r = 2$ and $\theta \in \frac {\pi}6 + \frac{2\pi}3\mathbb Z$

simply a way of saying

  1. $z = r cis \theta$ where $r = 2$ and $\theta \in \frac {\pi}6 + \frac{2\pi}3\mathbb Z$

which is the same as saying

  1. $z \in \{2cis\theta|\theta = \frac {\pi}6 + \frac{2k\pi}3; k\in \mathbb Z\}$

which is what the statement

  1. $z \in 2cis \frac {\pi}6 + \frac{2\pi}3\mathbb Z$

means?

So, is your question one of notation and how exactly do we define these set notations? Or is there some logical step along the way confusing you?

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  • $\begingroup$ I numbered your lines to facilitate discussion, hope that's okay. I think the bit that's really throwing me is the step from (2) to (3). These certainly aren't equivalent in the ordinary sense of the word; for instance, the former has free variables, whereas the latter does not. $\endgroup$ – goblin Nov 21 '16 at 9:06
  • $\begingroup$ @goblin: (2) is sloppy writing and actually has unwritten existential quantifiers around $r$ and $\theta$. Similarly, the set-builder notation in (3) secretly has the same existential quantifiers since $\{f(x):x\in A\}$ actually means $\{y:y=f(x)\text{ for some }x\in A\}$. When you write what these shorthand notations actually mean precisely with all the quantifiers, (2) has the form "$z$ satisfies property $P$" and (3) has the form "$z\in\{x:x\text{ satisfies property }P\}$", for the same $P$. $\endgroup$ – Eric Wofsey Nov 21 '16 at 9:19
  • $\begingroup$ @EricWofsey, dude, I think you've hit on it!!! There's implicit existential quantification going on that I just wasn't seeing. Do you mind if I make that an answer? $\endgroup$ – goblin Nov 21 '16 at 9:21
  • $\begingroup$ @goblin: No, but I already wrote such an answer myself. $\endgroup$ – Eric Wofsey Nov 21 '16 at 9:30
  • $\begingroup$ Well, it's nice to identify where the problem lies. I don't much do not like the notation among sets A+B to mean the set {a+b|a in A; b in B} but I admit it's useful. I really dislike the notation 5X to mean {5x|x in X} but many take it as obvious. In this case $\pi/6 + 2\pi/3*Z$ to mean {$y = \pi/6 + 2\pi k/3| k \in \mathbb Z$}$ seems very sloppy and abusive, but I thought it was clearly meant to be nothing more than shorthand notation. $\endgroup$ – fleablood Nov 21 '16 at 16:38
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All that's going on here is that both statements use slightly sloppy notation. When you make their notation precise, they are saying exactly the same thing.

First, the statement $$(r=2) \mathrel{\&} \left(\theta \in \frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right)$$ doesn't really make sense, because $r$ and $\theta$ are not well-defined (well, $r$ is, but $\theta$ isn't). What this statement actually means in more precise language is $$\text{there exists $\theta\in\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}$ such that $z=2\operatorname{cis}(\theta)$}.\quad (*)$$

Similarly, the statement $$z \in 2 \mathop{\mathrm{cis}}\left(\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right)$$ involves a bit of abuse of notation (though this abuse is fairly standard), applying operations on numbers to sets of numbers. That is, what this notation really means is that $z$ is in the image of the set $\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}$ under the map $\theta\mapsto2\operatorname{cis}(\theta)$. But if you write down the definition of this image, it is exactly the set of $z$ that satisfy $(*)$ above. So the two statements are equivalent.

(There are other ways you might interpret the first statement such that it would require a little more work to show it is equivalent to the second. For instance, you might interpret it as saying that $r(z)=2$ and $\theta(z)\in A$ where $r(z)=|z|$ and $\theta(z)$ is the argument of $z$ as an element of $\mathbb{R}/2\pi\mathbb{Z}$, and $A$ is the image of $\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}$ in $\mathbb{R}/2\pi\mathbb{Z}$. To prove this is equivalent to the existence of $\theta$ as in $(*)$, you need to use the fact that $z=r\operatorname{cis}(\theta)$ iff $r=r(z)$ and $\theta+2\pi\mathbb{Z}=\theta(z)$.)

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Then, with a bit of work, we can show:

$$z^3 = 8i \iff (r=2) \mathrel{\&} \left(\theta \in \frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right)$$

In my reading, the RHS translates to: $$z \in \Big\{\,r \mathop{\mathrm{cis}} (\theta) \in \mathbb{C} \;\big|\; r = 2 \Big\} \;\cap\; \Big\{\,r \mathop{\mathrm{cis}} (\theta) \in \mathbb{C} \;\big|\; \theta \in \left(\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right) \Big\}$$.

It follows that

$$z^3 = 8i \iff z \in 2 \mathop{\mathrm{cis}}\left(\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right),$$

Again in my reading, the RHS translates to: $$z \in \Big\{\,r \mathop{\mathrm{cis}} (\theta) \in \mathbb{C} \;\big|\; r = 2 \;\land\; \theta \in \left(\frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}\right) \Big\}$$

The two look equivalent to me as just a matter of notation, with no principles of logic involved.


As a side comment, I am not a big fan of either notation, since both obfuscate the fact that each set actually contains exactly $3$ elements.

The most direct (if slightly cheating) derivation of the result set could be rewriting the equation as: $$\left(\frac{i\,z}{2}\right)^3=1$$ with the well known cubic roots of unity as solutions $\frac{i\,z}{2} \in \{ 1, \omega, \omega^2 \mid \omega = \mathop{\mathrm{cis}} (\frac{2 \pi}{3}) \,\}$.

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    $\begingroup$ I will admit I'm not crazy about the notation of $2 \mathop{\mathrm {cis}} \frac {\pi}6 + \frac{2\pi}3\mathbb Z$ as set notation to mean the set $\{2\mathop{\mathrm {cis}} \frac {\pi}6 + \frac{2k}{\pi}|k \in \mathbb Z\}$ which I suspect may be causing a lot of the confusion, but it's not without precedence. $\endgroup$ – fleablood Nov 20 '16 at 17:09
  • $\begingroup$ The two aren't purely formally equivalent: it might be that $z=2\operatorname{cis}(\theta)$ for some $\theta$ and $z=r\operatorname{cis}(\theta)$ for some $\theta\in \frac{\pi}{6}+\frac{2\pi}{3}\mathbb{Z}$, but you can't choose $r$ and $\theta$ such that both are true at once. You need to use the fact that $r$ is unique, so if you have one representation with $r=2$ then every representation has $r=2$. $\endgroup$ – Eric Wofsey Nov 21 '16 at 9:16
  • $\begingroup$ @EricWofsey Unless I misread it, the intersection would be the empty set in that case using either notation. I still see it as nothing more than: $$\{(a,b) \in C \mid P(a)\} \cap \{(a,b) \in C \mid Q(b)\} \equiv \{(a,b) \in C \mid P(a) \land Q(b)\}$$ which would be considered a matter of notation in most contexts except maybe axiomatic logic. $\endgroup$ – dxiv Nov 21 '16 at 17:48
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The It follows that is merely saying that if the solutions form the set

$$\{(r_k,\theta_k)\}$$ in polar coordinates, the solutions in complex numbers are

$$\{z_k=r_k\text{ cis }\theta_k)\}.$$

This is essentially a restatement of how the polar coordinates work. I don't see what is "unrigorous" here.

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