We all know that [latex]e^{i \cdot \pi} = -1[/latex]. With some algebraic manipulation, we can get supposed paradoxes from this:

(1) [latex]e^{2\pi i} = 1[/latex]

(2) [latex]e^{2\pi i + 1} = e[/latex]

(3) [latex](e^{2\pi i + 1})^{2\pi i + 1} = e[/latex], substituting this into itself.

(4) [latex]e^{(2\pi i + 1)^{2}} = e[/latex]

(5) [latex]e^{-4\pi^{2} + 4\pi i + 1} = e[/latex]

(6) [latex]e^{-4\pi^{2}}e^{4\pi i} = 1[/latex]

(7) [latex]e^{-4\pi^{2}} = 1 = e^{0}[/latex]

(8) [latex]-4\pi^{2} = 0[/latex]

You might think that this problem arises because we used the exponent trick at (7) and (8), but this is true for real numbers (as the exponential is bijective on the reals). So, what went wrong? Look at step (3) and (4). We assumed we could just multiply those complex numbers down, as it works for the real numbers. But, as it turns out, it does not work for complex numbers. This is because:

[latex]e^{0} = e^{\pi i}[/latex]

[latex](e^{0})^{i} = (e^{\pi i})^{i}[/latex]

[latex]e^{0} = 1 = e^{-pi}[/latex]

So, this problem arises because the representation of a complex number as an exponentiation is not unique. But, if we assume that in [latex](a^{b})^{c}[/latex], [latex]c[/latex] is a positive real number, this works. This same problem arises in the logarithm, as

[latex]i\pi = log(-1) = log((-i)^{2}) \ne 2log(-i) = 2(-\frac{i\pi}{2}) = -i\pi[/latex]

This sort of failure happens a lot, like with [latex](bc)^{x} = b^{x}c^{x}[/latex] gives

[latex]1 = (-1 \cdot -1)^{\frac{1}{2}} \ne (-1)^{\frac{1}{2}} \cdot (-1)^{\frac{1}{2}} = -1[/latex]

Find some other paradoxes and rules that don't work with complex numbers, most of which will involve exponentiation. This just goes to show that you can't take rules for granted when changing what field you're in, like how some properties of certain groups don't transfer over to other groups (which I will make an article about how this relates to specific vector spaces and groups).

That one guy that's always on

i'll put this in an article. Put this in a latex converter if you're confused.

That one guy that's always on

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