0
votes
2answers
121 views

Why is this wrong (complex numbers and proving 1=-1)?

$$(e^{2πi})^{1/2}=1^{1/2}$$$$(e^{πi})=1$$ $$-1=1$$ I think it is due to not taking the principle value but please can someone explain why this is wrong in detial, thanks.
8
votes
0answers
107 views

Complex Exponential False “Proof” That All Integers Are $0$

The following false "proof" is attributed to Thomas Clausen in 1827, and was stated on page 79 of Nahin's An Imaginary Tale. $e^{i2\pi n}=1$ for all integers $n$. So \begin{align*} ee^{i2\pi ...
20
votes
9answers
3k views

Why is $i^3$ (the complex number “$i$”) equal to $-i$ instead of $i$? [duplicate]

$$i^3=iii=\sqrt{-1}\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i $$ Please take a look at the equation above. What am I doing wrong to understand $i^3 = i$, not $-i$?
18
votes
9answers
2k views

$1/i=i$. I must be wrong but why? [duplicate]

$$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \sqrt{\frac{1}{-1}} = \sqrt{-1} = i$$ I know this is wrong, but why? I often see people making simplifications such as ...
3
votes
1answer
190 views

Find the flaw in my proof that $z^2 =1$ has more than $2$ distinct solutions.

Let $z \in \mathbb{C}$ be any number that satisfies the equation $z^2=1$. Certainly, $z=\pm1$ are two possible solutions to this equation. I claim that $z^k$ is also a solution to this equation for ...
0
votes
1answer
84 views

Why is the second equality wrong?

Here's a "proof" of $e^x=1$ for all $x$: $$\exp(x)=\exp\left(i2π⋅\frac{x}{i2π}\right)=\bigl(\exp(i2π)\bigr)^{x/(i2π)}=1^{x/(i2π)}=1$$ Why is the second equality wrong?
25
votes
1answer
926 views

What is wrong with this fake proof $e^i = 1$?

$$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$ Obviously, one of my algebraic manipulations is not valid.
10
votes
3answers
582 views

Why this proof $0=1$ is wrong?(breakfast joke)

We have $$e^{2\pi i n}=1$$ So we have $$e^{2\pi in+1}=e$$ which implies $$(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$$ Thus we have $$e^{-4\pi^{2}n^{2}+4\pi in+1}=e$$ This implies ...
4
votes
1answer
542 views

What's wrong with this proof that $e^{i\theta} = e^{-i\theta}$?

I recently learned that $\cos{\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and $\sin{\theta} = \frac{e^{i\theta} - e^{-i\theta}}{2}$ Based on this, I managed to "prove" that: $$e^{i\theta} = ...
10
votes
4answers
543 views

Which step in this process allows me to erroneously conclude that $i = 1$

I was playing around with imaginary numbers and exponents and came up with this: $$ i = \sqrt{-1} $$ $$ \sqrt{-1} = (-1)^{1/2} $$ $$ (-1)^{1/2} = (-1)^{2/4} $$ $$ (-1)^{2/4} = ((-1)^{2})^{1/4} ...
1
vote
1answer
404 views

paradoxical answers using 'i' [duplicate]

Possible Duplicate: -1 is not 1, so where is the mistake? Significance of $\displaystyle\sqrt[n]{a^n} $? $i = \sqrt{-1} = \sqrt{\frac{-1}{1}} = \sqrt{\frac{1}{-1}} = \frac{1}{i}$ hence, ...
1
vote
3answers
339 views

A contradiction involving exponents

Where is the error in the following statement: $i^2=(i^2)^{\frac{4}{4}}=(i^4)^{\frac{2}{4}}=(1)^{\frac{1}{2}}=1$? I feel the error is in the first equality, because $(i^2)^{\frac{4}{4}}$ is in fact ...
26
votes
10answers
3k views

$i^2$ why is it $-1$ when you can show it is $1$? [duplicate]

We know $$i^2=-1 $$then why does this happen? $$ i^2 = \sqrt{-1}\times\sqrt{-1} $$ $$ =\sqrt{-1\times-1} $$ $$ =\sqrt{1} $$ $$ = 1 $$ EDIT: I see this has been dealt with before but at least with ...
27
votes
6answers
2k views

$-1$ is not $ 1$, so where is the mistake?

I know there must be something unmathematical in the following but I don't know where it is: \begin{align} \sqrt{-1} &= i \\ \\ \frac1{\sqrt{-1}} &= \frac1i \\ \\ \frac{\sqrt1}{\sqrt{-1}} ...