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The ln(i^2) one is actually kind of interesting. How would you address it?
6ln(i) = 2ln(i)
i^6 = i^2
iÂ²*iÂ²*iÂ² = iÂ²
-1 * -1 * -1 = -1
When a and b are real numbers, ln a^b = b ln a, but if not, you can’t garantee. e^ix = cos x + isen x, so e^ix = e^i(x+2pi). You can see that exponential is not one-to-one in the complex field, that’s why ln behaves so strangely.
e^ipi = -1
ln -1 = i(pi + 2kpi), k integer
If you let k = 1 and 3 at the same time
2 = 6
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