0

Possible Duplicate:
$i^2$ why is it $-1$ when you can show it is $1$?

Try to find what's wrong there:
nb: the squareroot can be defined for all complex numbers as $\exp(1/2\cdot\log(z))$

$$1 = \sqrt 1 = \sqrt{-1\cdot-1} = \sqrt{i^2\cdot i^2} = \sqrt{i}^2 \cdot\sqrt{i}^2 = i \cdot i = -1 $$

edit:
$ \sqrt{i} = \exp(1/2\cdot\log(i)) $

to find $\log(i)$:
$$\begin{align} &x+iy = \log(i)\\ &\exp(x)\exp(iy) = i\\ &\exp(x)\exp(iy) = i\\ &\exp(x)(\cos(y)+i\sin(y)) = i \end{align}$$ we arrive at $y=\pi/2$ and $x=0$

so $ \sqrt{i} = \exp(1/2\cdot i\cdot\pi/2) $

and $ \sqrt{i}^2 = i $

where are the multiple values?

Community
  • 1
caub
  • 267
  • 2
    Square root of complex number is multiple-valued – Norbert May 12 '12 at 22:02
  • could you reopen the question, the said duplicates are not – caub May 12 '12 at 22:33
  • The answer to why the derivation of $1=-1$ is invalid yet seemingly legitimate is the very topic of, and is exposed in the responses to, the linked question. Your edit includes some equations and "where are the multiple values?" -- is this supposed to be interpreted by others as a distinct question of sorts? Trying my best to make sense of it, I imagine you're looking for an adaptation of the original false proof to (invalidly) find multiple values again for $\sqrt{i}^2$. If so, you should always ask separate questions as separate questions. – anon May 12 '12 at 23:27
  • well had to search far in the answers, gary's answer seems good, tx – caub May 13 '12 at 06:39

0 Answers0