Rearrange Euler's identity to isolate i

Gondur
Messages
25
Reaction score
0

Homework Statement



Maybe this is not possible because i does not represent anything quantile and is merely abstract? I'm not sure and maybe you guys can help!


Homework Equations



e^{i \pi} + 1 = 0

The Attempt at a Solution



e^{i \pi} + 1 = 0

e^{i \pi} = -1

You cannot take natural log of a negative number so where do I go from here?

ln(e^{i \pi})=ln(-1)

i \pi=ln((-1))

i=\frac{ln(-1)}{\pi}
 
Physics news on Phys.org
You would have to take the complex logarithm, which is a subtle little thing.

https://www.physicsforums.com/showthread.php?t=637214
 
Last edited by a moderator:
It is true that ln(-1) = ipi. I don't see anything wrong with what you've said.

You cannot take the ln of a negative in the reals. We are explicitly not limited to the reals.
 
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