MedVision ad

Hellooooooo fellow 4 uniter's (1 Viewer)

b3kh1t

Member
Joined
May 21, 2010
Messages
271
Gender
Male
HSC
2011
Who reckons they know what is , yes that is i to the power of i :p
 
Last edited:

mirakon

nigga
Joined
Sep 18, 2009
Messages
4,221
Gender
Male
HSC
2011
hup is correct, its a real value

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

D94

New Member
Joined
Oct 5, 2011
Messages
4,423
Gender
Male
HSC
N/A
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

Member
Joined
Jun 7, 2010
Messages
177
Gender
Male
HSC
2011
? 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

Member
Joined
May 21, 2010
Messages
271
Gender
Male
HSC
2011
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

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Using the standard definition of complex powers as multi-valued functions:



where n ranges over all integers.
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007


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

 

mirakon

nigga
Joined
Sep 18, 2009
Messages
4,221
Gender
Male
HSC
2011
i do not think Euler's Identity is syllabus, but if you do a bit of extra-curricular maths you may know it
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top