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S
Isn't that what the i2kpi is for? Why can you add the second i2lpi when raise it to a complex number?
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S
How does the red part come into play $$2l\pi i$$?
$$((-i)^{(2+i)})^{(2-i)} = e^{((2-i)((2+i)(-i\pi/2 + 2k\pi i) +2l\pi i))}$$