3 ms·
The votey (red button to the bottom right of comic) says i^i is a real number. I didn't understand how at first, but the proof seems straightforward: https://ma
by oceliker 5y ago
The votey (red button to the bottom right of comic) says i^i is a real number. I didn't understand how at first, but the proof seems straightforward: https://math.stackexchange.com/questions/216871/why-is-ii-real https://math.stackexchange.com/questions/216871/why-is-ii-re...
- whatgoodisaroad 5y agoThat proof uses Euler's formula, which is the real brain melter.
- warent 5y agoI vaguely remember a 3blue1brown video that explains this really well too. Something to do with thinking about it in terms of rotations on the complex number plane which can land you back on the axis of real numbers. That was the most intuitive thing that I remember seeing
- nohuck13 5y agohttps://youtu.be/pq9LcwC7CoY https://youtu.be/pq9LcwC7CoY
- camjw 5y agoOverly pedantic but I think i^i is actually not well-defined (and in general exponentiation of complex numbers is not well defined) since e^(2 * pi * i) = 1 so for instance we would have i = e^(0.5 * pi * i) = e^(2.5 * pi * i) and then i^i = e ^(-0.5 * pi) = e^(-2.5 * i) but those last two numbers are both real and not equal so complex exponentiation only works when you pick a branch of the complex logarithm. Anyway...
- shonenknifefan1 5y agoYes, though regardless of which branch you pick to define i^i, it's value is real.