# Complex exponential function

1. Mar 16, 2007

### sara_87

good evening all!

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

Determine the exact values of
$$j^j$$

2. Relevant equations

j = sauare root of -1

3. The attempt at a solution

stuck

Last edited by a moderator: Jan 3, 2014
2. Mar 16, 2007

### Integral

Staff Emeritus
What do you know about the complex exponential function? Surely your text has some information you can apply.

3. Mar 16, 2007

### sara_87

i=-1^(1/2)

4. Mar 16, 2007

### sara_87

i'm still stuck

5. Mar 16, 2007

### sara_87

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?

6. Mar 16, 2007

### matt grime

YOu won't get the answer that way (since it doesn't deal with branches properly).

Complex numbers have representations as r*exp(jt) for r, j in R. Use it.

7. Mar 16, 2007

### sara_87

does exp(-2pi) = -1

8. Mar 16, 2007

### sara_87

no it doesn't

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

9. Mar 16, 2007

### matt grime

No. If you must, put it in your calculator, but e^{-2pi} is obviously a positive real number, and something in the region of 1/2^64.

10. Mar 16, 2007

### matt grime

solve what? exp(-2pi) is right. It just isn't anything nicer than that.

11. Mar 16, 2007

### sara_87

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

( $$j^j$$ <--that is almost the only thing i can write in latex form)

Last edited: Mar 16, 2007
12. Mar 16, 2007

### sara_87

yes it does

13. Mar 16, 2007

### matt grime

no it doesn't. j (or i) is exp(j*pi/2) so j^j is..... You can even google the phrase

i to the i

or

i^i

14. Mar 16, 2007

### sara_87

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

15. Mar 16, 2007

### matt grime

Well keep trying 'til you do get zero. x/(e^x-1) differentiate it three times... eugh, no thanks. Do it the smarter way - sub in e^x first, then do the 1/(1-s) expansion. collect terms and you get

-1/4! + 2/2!3! - (1/2!)^3 = -1/24+1/6-1/8=0

(and when did 0 cease to be a number?)

16. Mar 16, 2007

### tim_lou

ha.

let
$$j=e^{jx}$$
$$j^j=e^{j\cdot jx}=e^{-x}$$

so what should be the answer? (certainly not -1)

Last edited: Mar 16, 2007
17. Mar 16, 2007

### sara_87

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

$$e^x$$ = 1 + x + $$x^2$$/2! + $$x^3$$/3! + ...

$$e^x$$ - 1 = x + $$x^2$$/2! + $$x^3$$/3! + ...

x/($$e^x$$ - 1) = x/(x + $$x^2$$/2! + $$x^3$$/3! + ...)

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

(1+ x/2! + $$x^2$$/3! + $$x^3$$/4! +...)^(-1)

let x/2! + $$x^2$$/3! + $$x^3$$/4! +.. = X

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

so (1+ x/2! + $$x^2$$/3! + $$x^3$$/4! +...)^(-1) = 1 - x/2! - $$x^2$$/3! - $$x^3$$/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???

Last edited: Mar 16, 2007
18. Mar 16, 2007

### sara_87

yes, okay, cheers, i get the picture

(<-- this is my favourite smiley)

19. Mar 16, 2007

### matt grime

Why does B_2=-1/3? What happend to the higher powers of X (a bad choice. Y would be better). You have only take the contribution from X^1, not X^2 or higher.

and 1/(1+s) is 1-s+s^2-s^3+s^4+...

Now put s = x/2!+x^2/3!+x^3/4!+...

Last edited: Mar 16, 2007
20. Mar 16, 2007

### sara_87

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!!!