Question

Simplify:

e^(i(1-n)pi)

Answer 1

\[e ^{i(1-n)\pi}\]

and

\[e ^{i(1+n)\pi}\]

Answer 2

I know that

\[e ^{i \pi}=(-1)^{n}\]

Answer 3

and what about
\[e ^{-i n \pi}=(-1)^{n}?\]

Answer 4

\(e^{in\pi}=(-1)^n\) for \(n=1,2,3,..\).

Answer 5

You can get from that \(e^{i\pi}=-1\).

Answer 6

I wrote that above.

Answer 7

Read his response properly, what you said about e^i*Pi is not correct.

Answer 8

oops, but

i have this:

\[e ^{(1-n)\pi}=e ^{i \pi}e ^{-i n \pi}\]
\[e ^{i \pi}e ^{-i n \pi}=(-1)^{n}(-1)^{n}=(-1)^{2n}=1???\]

Answer 9

i forgot the "i"

Answer 10

This is not true!

Answer 11

\(e^{i(1-n)\pi}=e^{i\pi}e^{-in\pi}=-1(-1)^n=(-1)^{n+1}\), where \(n=1,2,3,..\).

Answer 12

\[e^{i*\pi} = -1 , jessica\] that was your problem, you were thinking it was \[(-1)^n\] as well

Answer 13

In other words you will have -1 for n even and 1 for n odd.

Answer 14

owhhh sh*t! e^(i*pi) is -1 =.= and e^(i*n*pi)=(-1)^n

Answer 15

stupid me!

Answer 16

Haha finally :P

Answer 17

thank you guys <3

Answer 18

You're welcome.

Answer 19

No problem =)