Question

What is the value of e^-i(pi/3) ?
Please see the link below for better understanding:
http://imageshack.us/photo/my-images/24/70174469.png/
Thanks

Answer 1

I don't like complex numbers... they're so... complex :D
But, we can't live without them :)
Use this identity...

\[\huge e^{i\theta}=\cos \ \theta + i \ \sin \ \theta \]
Noting that theta must be in radians :)

Answer 2

Thank you very much

Answer 3

Anytime :)

Answer 4

Just one more question, in my question I have -ve sign in power of e (e^-iota theta), does that make difference ?

Answer 5

Well, you have
\[\huge e^{-i\left(\frac{\pi}{3} \right)}\]
Which is really just the same as
\[\huge e^{i\left(-\frac{\pi}{3} \right)}\]

Answer 6

Does that answer your question? :)

Answer 7

so it means that: e^-i(pi/3)=cos (-pi/3)+i sin (-pi/3)

Answer 8

Nicely done :)

Answer 9

and can we apply de moivre's theorem in anyway to it, because my final answer should be 1.

Answer 10

Really? Hang on...

Answer 11

I don't think it's 1.
\[\huge e^{i (2\pi)} = 1\]
but not sure about pi/3

Answer 12

ok , thanks you helped a lot.

Answer 13

You do understand why
\[\huge e^{i (2\pi)} = 1\]
right?
cos 2pi + isin 2pi = 1 + 0
:)

Answer 14

sin2pi=0, then what happend to iota how you got isin2pi=0 ? simple multiplication of i with 0?

Answer 15

Yes. i times 0 is 0, just like with real numbers.

Answer 16

so in my case what will be the final answer: cos(pi/3)+isin(p1/3)
cos(pi/3)= 1/2 and sin(pi/3)=sqrt(3)/2
which makes final answer as 1/2+isqrt(3)/2 ?

Answer 17

Yeah, I'm afraid so...
How were you supposed to get 1?
:/

Answer 18

but I supposed to get rid of imaginary part (iota) anyway thanks maybe I made a mistake before.

Answer 19

:)

Answer 20

please have a look at this