Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $z \in \mathbb{C}$ be any number that satisfies the equation $z^2=1$. Certainly, $z=\pm1$ are two possible solutions to this equation. I claim that $z^k$ is also a solution to this equation for any $k \in \mathbb{R}$, resulting in (probably) at least one other distinct solution (for example, the solution $z=(-1)^{\pi}$).

Proof: Choose any $k \in \mathbb{R}$. Then $(z^k)^2=z^{2k}=(z^2)^k=1^k=1$, as desired.

So, can someone tell me what's wrong with my proof? I know that this has something to do with roots of unity and that this equation should only have 2 distinct solutions, so something must be wrong with my proof. Please bear in mind that I understand very little about complex numbers besides the definition that $i^2 = -1$. Thanks!

share|cite|improve this question
Once you allow $z$ to be a complex number, it is no longer true in general that $z^{2k}=(z^2)^k$. Take a look at this old answer of mine. – Zev Chonoles May 23 '13 at 6:14
Another way to look at this is that $z^k$ can be interpreted in more than one way when $k$ is a non-integer. If you think about it, you're really taking $\exp(k \log z)$, and $\log z$ can also take multiple values in $\mathbb C$ (think $e^{2\pi i} = e^0$). – Erick Wong May 23 '13 at 6:21
up vote 4 down vote accepted


The rule $\sqrt{x} \sqrt{y}=\sqrt{xy}$ is generally valid only if both $x,y \in +ve \text { real numbers}$

Look at one more:

$$e^{i 2\pi}=1$$

$$(e^{2 \pi i})^i=1^i$$

$$e^{-2 \pi}=1$$

The error here is that the rule of multiplying exponents as when going to the third line does not apply unmodified with complex exponents, even if when putting both sides to the power i only the principal value is chosen.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.