# Tagged Questions

98 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.
90 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 ...
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 $-i$?
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 ...
168 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 ...
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?
881 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.
559 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 ...
499 views

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: $$e^{i\theta} = ... 4answers 531 views ### Which step in this process allows me to erroneously conclude that i = 1 I was playing around with imaginary numbers and exponents and came up with this:$$ i = \sqrt{-1}  \sqrt{-1} = (-1)^{1/2}  (-1)^{1/2} = (-1)^{2/4}  (-1)^{2/4} = ((-1)^{2})^{1/4} ...
371 views

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, ...
320 views

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 ...
### $i^2$ 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 ...