Obvious Theorems (Which Are False)

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[Blaqjaqshellaq]

In fact, i^i has an infinite set of real solutions!

[OptimusPhillip]

Another way to approach this: by Euler's identity, it can be shown that i=exp(i*pi/2). Raise both sides to the power of i, and you get i^i=exp(i*i*pi/2), or i^i=exp(-pi/2)

At least, that's the principle value. There are infinitely many others, due to the circular nature of Euler's identity.

[erikkonstas]

TBF real numbers can be expressed as complex numbers as well.

[FrancoRe-o]

The thing is that you cant use the real number propiety for logarithms, ln(i²) isnt the same as 2ln(i)

[leesweets]

No one makes the claim that combining imaginary numbers makes imaginary numbers. Simply adding 3i to -3i, or multiplying i with i, are both simple, trivial counterexamples. The very definition of i is a counter example; you cant employ i without implicitly acknowledging the claim is untrue.

[ishayisrael]

Of course this theorem isn't true, and you don't have to go to i^i to see that. Take for example i*i which is -1

[AdamReveland]

Since the complex numbers form a field, any combination of two non zero complex numbers will always result in a complex number.

[sankoshroy]

How can you use logarithmic function here 🤔 This function only contains real positive numbers in its domain . Yes the result is true and we can prove this using Euler's identity but how can you use this

[leytonbennet]

The complex numbers are a closed set though; all real numbers are complex as well.

[emmecuh]

It's true though, considering real numbers are a subset of imaginary numbers.

[atheybengala]

Well, i mean, e is a complex number too