# Why is this wrong (complex numbers and proving 1=-1)?

$$(e^{2πi})^{1/2}=1^{1/2}$$$$(e^{πi})=1$$ $$-1=1$$ I think it is due to not taking the principle value but please can someone explain why this is wrong in detial, thanks.

• Since this appears in your question, what is your definition of $x^{1/2}$ when $x$ is not a nonnegative real number? – Did Jun 21 '14 at 12:50
• Note that "proof" is a thing: a noun. "Prove" is the corresponding verb: We aim to prove that X holds by writing a proof that X holds. – Namaste Jun 21 '14 at 12:59
• sorry i am dyslexic and hence why there is often a lot of grammar + spelling mistakes in my questions! – user135842 Jun 21 '14 at 13:00
• possible duplicate of $-1$ is not $1$, so where is the mistake? – Jyrki Lahtonen Jun 21 '14 at 13:26

You used two different branches of the function $x^{\frac{1}{2}}$.
Note that even in exponential form $(e^{x})^\frac{1}{2}$ has two different branches: $e^{\frac{x}{2}}$ and $e^{\frac{x}{2}+\frac{2 \pi i}{2}}$.
$$\sqrt{(-1)^2} = \sqrt{(1)^2} \quad \Rightarrow \quad -1 = 1$$
but the square root is not a map, i.e. $\sqrt{a}$ can have multiple solutions. In particular, it is not the inverse of $x^2$.