Complex exponential function

good evening all!

1. The problem statement, all variables and given/known data

Determine the exact values of
[tex]j^j[/tex]

2. Relevant equations

j = sauare root of -1

3. The attempt at a solution

stuck

 

i=-1^(1/2)

 

i'm still stuck

 

okay this is what i did:

((-1)^(1/2))^((-1)^(1/2))

i know this looks crazy but i still don't know how to use latex properly ...if you copy it out on a paper you'll understand what i'm trying to write...then i multiplied -1 with the 1/2...i can do that right?

 

does exp(-2pi) = -1

 

no it doesn't




wait no one give me any tips i think i can solve it

 

okay...i was about to say i'm stuck again but now i know that my teacher is crazy...but just to make sure [tex]j^j[/tex] doesn't equal -1...does it?

( [tex]j^j[/tex] <--that is almost the only thing i can write in latex form)

 

yes it does

 

my teacher is very very clever man and he said that the answer is -1 but he didn't use any of the complex number rules...he did something else! and he got -1

sorry i used this formula to show that it does equal -1:

r[cos(O) + jsin(O)] = rexp(jO)

and in my desperate want to get this question right i didn't see the j next to the O, and so after i substituted the numbers i'm obviously going to get -1...never mind!

( O = theta)

using your advice i went on google and thank god for wikipedia i got the answer...and i understand this time (for a change)

you know i'm still stuck on that bernoulli number question...for B_3 you're supposed to get zero i actually got a number

 

i took your advice ( i kinda knew it anyway) but i'm going to write out in full what i did:

[tex]e^x[/tex] = 1 + x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...

[tex]e^x[/tex] - 1 = x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...

x/([tex]e^x[/tex] - 1) = x/(x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...)

divid top and bottom by x => 1/(1 + x/2! + [tex]x^2[/tex]/3! + ...)

(1+ x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +...)^(-1)

let x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +.. = X

(1 + X)^(-1) = 1- X + [((-1)(-2))/2!] [tex]X^2[/tex] +...

so (1+ x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +...)^(-1) = 1 - x/2! - [tex]x^2[/tex]/3! - [tex]x^3[/tex]/3! -...((-1)(-2))/2!(X)^2 +...

so B_1 =-1/2, B_2= -1/3!, B_3 = -1/4!.

B_2 and B_3 are obviously wrong but i don't know where i'm going wrong???

 

yes, okay, cheers, i get the picture



(<-- this is my favourite smiley)

 

i still get a negative for B_2 i know what the answer is uppossed to be but i just can't see how...if i don't see why i don't sleep well!

i'm tired, you know i'm just going to live with the fact that i'm never going to get this question...

the thing is there is a minus outside the s and s^2 and the rest don't matter...

nevermind

thank you very much for your time and help!!!