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

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3. 1^-1^1/2 .... in short, it's imaginary

4. 1=e^(0)
1^i = (e^(0))^i = e^(0*i) = e^(0)=1

5. = 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.

6. 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.

7. Originally Posted by LeavingQuietly
= 1^(-1/2)

1^x = 1.

Originally Posted by Zelos
1=e^(0)
1^i = (e^(0))^i = e^(0*i) = e^(0)=1
I agreed with you :-D

8. I used
a<sup>b</sup> = e<sup>b ln(a)</sup>

=> 1

cheers

9. 1 to the power of anything is 1. 1 is just an identity element, remember?

1^bananas = 1

10. Originally Posted by dmerthe
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.

11. 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

12. 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>

13. yep soz i made that mistake i'll try again
thnx for pointing it out its a highschool maths thing that stupid me

14. Originally Posted by axlarry
Originally Posted by LeavingQuietly
= 1^(-1/2)

1^x = 1.

Originally Posted by Zelos
1=e^(0)
1^i = (e^(0))^i = e^(0*i) = e^(0)=1
I agreed with you :-D
I was wrong, that is how I explain it.

15. 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

16. Jaco - you taking logs in the complex plane and treating them as single valued functions instead of multivalued functions.

17. 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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