Why Does \(\frac{1}{i}\) Equal Both \(i\) and \(-i\)?

  • Thread starter Thread starter latentcorpse
  • Start date Start date
  • Tags Tags
    Confused
latentcorpse
Messages
1,411
Reaction score
0
i've confused myself

\frac{1}{i}=-i by multiplying by i/i

but then (\frac{1}{i})^2=-1 \Rightarrow \frac{1}{i}= i,-i
and clearly the positive root here doesn't make any sense?
 
Physics news on Phys.org
You are correct but you forgot to take the negative root of the left hand side.
 
Well you've somewhat answered your own question. You simply did pick the incorrect root as there are always 2 roots to a square root

\[<br /> \begin{array}{l}<br /> 3 = 3 \\ <br /> 3^2 = 3^2 \\ <br /> 9 = 9 \\ <br /> \sqrt 9 = \sqrt 9 \\ <br /> 3 = - 3? \\ <br /> \end{array}<br /> \]<br />

for example.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top