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Imaginary power (Posted on 2024-10-30)

Evaluate and comment:

f = i^i

No Solution Yet Submitted by Ady TZIDON

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soln

Comment 4 of 4 |

Just as 2^(1/2) is multivalued, i^i is multivalued.

i = e^ (i theta)

where theta = pi/2, pi/2 +/- 2 pi, pi/2 +/- 4pi, pi/2 +/- 6 pi, ...

So, i^i = e^[-( pi/2 + 2n pi)], n=any integer

Edited on November 12, 2024, 6:36 pm

Posted by Steven Lord on 2024-11-12 17:30:42

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