Question

Could someone explain this?

e^(2*S1*L)= -1
S1=((2n+1)*pi*i)/(2L)

Answer 1

\[e ^{2 * s1*L}=-1\]

\[s1 = ((2n+1)\pi*i)/(2L)\]

Answer 2

put s1 in the given equation
e^(2*((2n+1)pi*i)/2L*L)
let's just simplify the term in power
we have
2*((2n+1)pi*i)*L/2L
2L in numerator will get cancle with 2L in denominator , we 're left with
(2n+1)pi*i
if we put n=0 we het pi*i , if we put n=1 we get 3pi*i and so on
we e^((2n+1)i)
we know
\[e^(\theta*i)=\cos \theta+i \sin \theta\]
we know cos pi=-1 and sin pi=0
and so on cos 3pi=-1 sin 3pi=0
therfore our equation reduces to
cos (2n+1)pi+isin (2n+1)pi
cos(2n+1)pi=-1
sin(2n+1)pi=0
therfore we get
-1

Answer 3

uehhh, but what is you didnt know the answer... of S1

how would you solve it for S1?

Answer 4

i'm not using the answer, i'm putting s1's value in equation and then reducing it.
it's a standard equation
cos(2n+1)pi=-1
sin(2n+1)pi=0

Answer 5

oh you mean we have to find s1
sorry
that's also
easy

Answer 6

yeahh i mean... how can i find s1? and get the answer what i have typed above :)

Answer 7

we are given that e^(2*S1*L)=-1
that means that term in power is an odd multiple of pi
so
2*S1*L=(2n+1)pi
so
S1=(2n+1)pi/2L

Answer 8

where n=0,1,2,3

Answer 9

yeah but why is it odd?

Answer 10

how do you see that this equation is odd?

Answer 11

because odd multiple of pi gives
cos pi=-1
and sin pi=0

Answer 12

as there is no i term in the result , that means sin term is 0
which can only happen when theta = 0,pi or 2pi
but we want -1
cos pi , cos 3pi , cos5pi=-1
only odd multiple result in -1
therfore in general the theta is (2n+1)pi where n=0,1,2
if u put n=0 we get pi , n=1 3pi and so on

Answer 13

n can also be negative

Answer 14

you typed:

we are given that e^(2*S1*L)=-1 that means that term in power is an odd multiple of pi so 2*S1*L=(2n+1)pi so S1=(2n+1)pi/2L

The answer is S1=(2n+1)pi/2L * i

Answer 15

where did your "i" go

Answer 16

i will come

Answer 17

only the term other than i is equal to odd multiple of pi

Answer 18

i missed it but e^i*theta =cos theta +i sin theta

Answer 19

i still dont get is :(

Answer 20

could someone draw it out? i'm sure it has to do something with the unit circle

Answer 21

\[ e^z = -1 \]
if and only if
\[ z = \pi i + 2\pi k i, \ \ k \in \mathbb{Z} \]

Do you understand this much?

Answer 22

hi james,

no not really :s, but it consist only the Im. component

Answer 23

If we write z = a + bi, then \[ e^z \] equals what?

Answer 24

where a and b are real

Answer 25

a is the Re
b is the Im

Answer 26

of z. Yes. Now what is e^z equal to?

Answer 27

its must lie in the unit circle so i guess +1 or -1

Answer 28

ok. What does \[ e^{i \theta } \] equal?

Answer 29

i really dont know james...

Answer 30

For what value of theta does e^(i . theta) = 1 ?

Answer 31

owhh...

if theta is 0 or pi/2 = 1
if theta is pi or 3/2pi = -1

am i right?