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b3kh1t

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Who reckons they know what is , yes that is i to the power of i :p
 
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mirakon

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hup is correct, its a real value

about 0.207 is what e^(-pi/2) actually works out to
 

D94

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lol how when that will give you a real value hahaha

well Mirakon what you reckon??
Essentially, you need to know Euler's Identity to solve this.

e^(i.pi) + 1 = 0
e^(i.pi) = -1
e^(i.pi/2) = i (square root)
e^(-pi/2) = i^i (raise to the i; i*i = -1)
 

K4M1N3

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? if you have n=2 then it becomes e^(-pi) which I'm pretty sure is not equal to i^i (correct me if I am wrong)
buddy learn to read, n has a restriction for odd integers only!
 

b3kh1t

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Essentially, you need to know Euler's Identity to solve this.

e^(i.pi) + 1 = 0
e^(i.pi) = -1
e^(i.pi/2) = i (square root)
e^(-pi/2) = i^i (raise to the i; i*i = -1)
aiight but in the last line how did you get that
 

seanieg89

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Using the standard definition of complex powers as multi-valued functions:



where n ranges over all integers.
 

seanieg89

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It is basically just the usual index laws combined with the fact that:

 

mirakon

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i do not think Euler's Identity is syllabus, but if you do a bit of extra-curricular maths you may know it
 

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