Problematic Proof? (II)

Algebra Level 5

It is well known that $z^{n}=1$ (where n is an integer) has $n$ solutions evenly distributed around the unit circle. These are known as the $n^\text{th}$ roots of unity.

What about when $n$ is not an integer? What if $n$ is irrational? Here is a proof that $z^{r}=1$ has an infinite number of solutions when $r$ is irrational. There is also a conclusion as to what that means. Can you spot a flaw?

Step:

$1=e^{2ki\pi}$ for any integer $k$ so $z^{r}=e^{2ki\pi}$
$z=e^{2ki\pi/r}$
This argument will never be identical to a previous one as $k_{1}/r\neq k_{2}/r\text{ mod 1}$ for $k_{1}\neq k_{2}$
Since the number $z$ could have an infinite number of distinct arguments there must be an infinite number of numbers that satisfy the equation.

Conclusion:

Every number of magnitude 1 is a solution for $z$ to the equation $z^{r}=1$ , $r$ is irrational.

Step 1 is flawed Step 4 is flawed Steps 1 to 4 are correct, but the conclution is wrong. The proof and conclusion are correct Step 2 is flawed Step 3 is flawed

2 solutions

Mark Hennings
Mar 8, 2017

The attempted proof is incorrectly defining non-integer powers of complex numbers. The non-integer power of a complex number $z$ cannot be calculated for all $z \in \mathbb{C}$ at a time. We define $z^u \; = \; e^{u \log z}$ and we try $\log z = \log|z| + i\mathrm{Arg}\,z \hspace{2cm} z \neq 0$ The problem lies with the argument, since it is not uniquely defined on the whole complex plane. We need to make a conventional choice, such as demanding that $-\pi < \mathrm{Arg}\,z \le \pi$ for all $z$ . Other choices could be possible. It is important that a choice be made since, otherwise, the power $z^u$ is multivalued. The numbers $e^{iu\theta}$ and $e^{iu\theta + 2\pi iu}$ are distinct (so long as $u$ is not an integer), and these are equally good candidates for the $u$ th power of $e^{i\theta}$ . A convention about the argument is necessary so that we know which of the (in fact infinitely many) candidate values for $z^u$ to choose.

The most common convention is use what is called the principal branch of the argument . This causes us to remove the non-positive reals from the complex plane, and consider the set (called the cut plane ) $\mathbb{C} \backslash (-\infty,0]$ . Every $z$ in this set has a unique representation in the form $z \; = \; re^{i\theta} \hspace{2cm} r > 0\;,\; -\pi < \theta < \pi$ We can then define $\log z \; = \; \ln r + i\theta$ and hence define $z^u$ for any complex $u$ , but only for complex numbers in the cut plane .

It just so happens that this care is not necessary when $u$ is an integer.

Before we can talk about a function like $z^r$ where $r$ is irrational, therefore, we need to choose an appropriate branch of the argument. Suppose for the sake of argument that we have chosen the principal branch of the argument. The problem then arises at $\boxed{\mathrm{Step\; 2}}$ . We can only go from Step 1 to Step 2 by saying that $z = e^{\frac{2\pi ik}{r}}$ for such integers $k$ for which $\big|\tfrac{2k}{r}\big| < 1$ , so that $|k| < \tfrac12|r|$ .

To see why, suppose that $k$ was an integer with $|k| \ge \tfrac12|r|$ , and consider the complex number $z = e^{\frac{2\pi ik}{r}}$ . To calculate $z^r$ , we would have to find an integer $m$ such that $-\tfrac12 < \tfrac{k}{r} - m < \tfrac12$ , and regard $z$ in the form $z = e^{2\pi i(\frac{k}{r}-m)}$ with $-\pi < 2\pi(\tfrac{k}{r} - m) < \pi$ , so that we could then define $z^r$ , but then $z^r \; = \; e^{2\pi i(k- mr)} \; = \; e^{-2\pi imr} \neq 1$

If we chose a different branch of the argument, we would get a different range of appropriate $k$ for any $r$ , but there would still only be a finite number of them.

I believe I have an alternate way of dealing with the problem that does not require defining a branch of a logarithm. (I should clarify first however that the nonpositive reals are not "removed" from the complex plane when defining the principal logarithm, as you claim. It is simply the case that it is discontinuous (and, hence, not analytic) along the branch cut. At least this is the case from what I've seen. You may want to correct that...) I claim that step 2 can be corrected by simply adding 'there exists an integer k such that'. Here is my argument. I am aware that there are some missing details, but anyone can feel free to fill them in if they like. Also, feel free to alert me if you think there is a mistake. And if anyone doesn't think $r$ can be multiplied directly by the exponent in the way in which I do it, then please provide a counterexample that proves my method is incorrect. Otherwise, I don't wish to prove it's valid... But I certainly would not frown upon you proving it valid yourself. I'm just simply aware that it produces no error in the result. Ok, to the fun... Since $z^r = 1$ , $z$ is a unit complex number. Hence, let $t$ be a real number such that $z=e^{it}$ . Considering all angles, x=A, coterminal to t gives all representations of $z$ in the form $e^{ix}$ . So, trivially, $\bigvee_{n\in \mathbb{Z}} z = e^{i(t+2n\pi)}$ . Raising both sides to the $r$ power in each equation, and noting $z^r=1=e^{2\pi i k}$ for any integer $k$ , I get $\bigvee_{n,k\in \mathbb{Z}} e^{ir(t+2n\pi)}=e^{2\pi i k}$ , which implies $\bigvee_{n,m,k\in \mathbb{Z}} ir(t+2n\pi)+2\pi i m=2\pi i k$ . But since I can subtract $2\pi i m$ on both sides to obtain $2\pi i (k-m)$ on the RHS, and k - m is an integer, this is equivalent to (also dividing by $r$ ) $\bigvee_{n,k\in \mathbb{Z}} i(t+2n\pi)=\frac{2\pi i k}{r}$ . Exponentiating both sides leads to $z=e^{\frac{2\pi i k}{r}}$ for some k. Cheers? (I realize the $2n\pi$ was not necessary but I included it in there anyway.)