# Find the flaw in my proof that $z^2 =1$ has more than $2$ distinct solutions.

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!

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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

## 1 Answer

$$1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i^2=-1$$

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.

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