160 views

### Two contradictory ways to calculate $(e^{2\pi i})^i$? [duplicate]

As we know: $$e^{i2π} = 1$$ so here's the first way that we can calculate the expression in the title: $$(e^{i2π})^i = 1^i = 1$$ however, if before we simplify $e^{i2π}$ to $1$ we multiply the ...
114 views

### 1 = e^2π. Where did I make a mistake? [duplicate]

I seem to have proven that $e^{2\pi} = 1$. What is my mistake? See here for proof.
113 views

### Does $1 = 1 + 2i\pi$? [duplicate]

I was playing around with a method of how to do negative logs and cane up with the following method using Euler's identity: $$e^{i\pi} = -1$$ $$\therefore \ln{-1} = i\pi$$ So therefore the ...
81 views

### Rigorous proof of the argument that one cannot square $-1$ in complex number problems [duplicate]

I would like to ask a fundamental question with regards to the imaginary number and it is something many beginners are told are wrong, but I would like to seek a rigorous proof of why it is wrong. It ...
2k views

### Unexpected result from Euler's formula

I am a bit confused with a result I get from Euler's formula: $e^{2\pi i} = 1$ $\sqrt[3] { e^{2\pi i} }= \sqrt[3]{ 1 }$ $(e^{2\pi i})^{\frac{1}{3}}= 1$ $e^{\frac{2}{3} \pi i} = 1$ This last ...
148 views

### Is $y={(-1)^{x\overπ}+(-1)^{-x\overπ}\over 2}$ the same as $y=\cos x$?

I have always been intrigued by the equation $y=(-1)^x$, perhaps because it is so simple yet so difficult to find information about. It's the closest thing to a trigonometric function attainable using ...
### When is $(a^b)^c$ = $a^{bc}$ true?
I know that in some cases this rule is not true. For example $$((-1)^2)^\frac{1}{2}\ne(-1)^{(2\cdot\frac{1}{2})}$$ So when is this rule true ?