Linked Questions

3
votes
2answers
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 ...
0
votes
1answer
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.
-2
votes
1answer
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 ...
2
votes
1answer
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 ...
10
votes
5answers
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 ...
12
votes
2answers
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 ...
8
votes
3answers
166 views

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 ?
0
votes
3answers
179 views

Is $e^i =1$ a wrong belief in mathematics or it is true as shown below? [closed]

It is well known that $e^i$ is transcdental number and $| e^i|=1 $ , but if i will try to write $e^i$ by other way : Short Proof: $e^i =[(e^i)^{2\pi)}]^\frac{1}{2\pi}=[{{e}^{2i\pi}}]^{\frac{1}{...
0
votes
1answer
112 views

Complex exponentiation “proof” that $\pi = 0$ [closed]

$$\begin{align} e^{2 \pi i } &= 1 \\ e^{1 + 2 \pi i} &= e \\ e^{1 - 2 \pi i} &= e \\ {(e^{1 - 2 \pi i})}^{1 + 2 \pi i} &= e^{1 + 2 \pi i}=e\\ e^{1+4 \pi^2} &= e\\ e^{4 \pi^2} &...
4
votes
2answers
155 views

$n$-th root for a negative $n$

The $n$-th root of a number $x$, where $n$ is a positive integer, is a number $r$ which, when raised to the power $n$ yields $x$. [Wikipedia.org] But I don't see any problem with the following ...
4
votes
2answers
87 views

Why can't we separate fraction powers in euler's formula?

I am sure there is something wrong i am doing but i just want someone to point it to me. Why can't we say that $e^{\frac {\pi i}{3}} = \left(e^{\pi i}\right)^{\frac {1}{3}} = (-1)^{\frac {1}{3}} = -1$...
0
votes
1answer
100 views

What is solution of $j^3$ (j is complex number)?

I have a confused with this problem? I calculate this by 2 ways: $$j^3 = jj^2 = j(-1) = -j$$ $$j^3 = j^{\frac{12}{4}} = (j^{12})^{0.25} = 1^{0.25} = 1$$ Why does it have different result?