If the complex number i is defined as i=sqrt(1), then what is the value of 1^i?

If the complex number i is defined as i=sqrt(1), then what is the value of 1^i?
1^1^1/2 .... in short, it's imaginary
1=e^(0)
1^i = (e^(0))^i = e^(0*i) = e^(0)=1
= 1^(1/2)
1^x = 1.
*This is wrong, i thought i had the exponent rules in my head and all worked out. I didn't.
Depends on your branch of logarithm, 1^i is multivalued.
arg(1)=2*pi*n
for any integer n. So log(1)=2*pi*n*i, the choice of n corresponds to a choice of the bracnh of arg (equivalently a choice of branch of log), and a corresponding branch of z^i.
1^i=exp(i*log(1))=exp(i*2*pi*n*i)=exp(2*pi*n)
depending on which branch you are on.
though some texts will adopt the convention that a positive real base corresponds to the choice of arg that has arg of a positive real equaling zero.
I don't understand your answer, can you explain that again? ThanksOriginally Posted by LeavingQuietly
I agreed with you :DOriginally Posted by Zelos
I used
a<sup>b</sup> = e<sup>b ln(a)</sup>
=> 1
cheers
1 to the power of anything is 1. 1 is just an identity element, remember?
1^bananas = 1
Almost, 1 to the power of any real anything is always 1. Exponents are not that simple for complex numbers (as Schmoe mentioned) as the logarithm is multivalued.Originally Posted by dmerthe
i got + or  1
let x = 1^i
x = 1 ^ (1^(1/2))
x^2 = 1 ^ (1)
x^2 = 1/1 = 1
x = +/ 1
I'm not sure if it is right though
so yeh just check it to be sure but i know mathematically i didnt make mistakes its just i dont know if the negative should be part of the answer
Jako, you cant just square equations with out checking the answers  more importantly squaring the equation does not get rid of the problem in the exponent. You should have
x^2 = 1<sup>2 i</sup>
yep soz i made that mistake i'll try again
thnx for pointing it out its a highschool maths thing that stupid me
I was wrong, that is how I explain it.Originally Posted by axlarry
It is just plain 1
let x= 1^i
ln(x) = ln(1^i) ln on both sides
ln(x) =i * ln(1) ln(a^b)=b * ln(a)
e^ln(x) = e^[i*ln(1)] e on both sides
x = [e^i]^ln1 a^(b*c)= (a^b)^c
x =(e^i)^0 ln(1) = 0
x =1 a^0 = 1
Jaco  you taking logs in the complex plane and treating them as single valued functions instead of multivalued functions.
so i gotto go and try it again
this is good i love it when people keep on showing me just my mistakes and i gotto go and find the correction myself
thanks river rat
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