Question

someoen helpp pls.. medals and fan
question in comment

Answer 1

@IrishBoy123

Answer 2

@mathmate

Answer 3

@phi

Answer 4

my first thought is write W as
\[ w= \exp(i \ 2 \theta) = \cos(2 \theta) + i \sin(2 \theta) \]
and use the double angle identities
sin 2x = 2 sinx cosx
cos 2x = cos^2 x - sin^2 x

Have you tried that ?

Answer 5

yes tried it..

Answer 6

i dont know how to use it well.. so can you help in proving ?

Answer 7

First, do the top w-1
replace w with exp(i 2 theta) and then use Euler to rewrite in terms of cos and sin
can you do that ?

Answer 8

ive done all that.. and even replaced it in the above eqn

Answer 9

what do you get ?

Answer 10

\[(\cos 2\theta +i \sin \theta - 1)/ (\cos 2\theta + isin \theta +1 )\]

Answer 11

I assume you mean i sin 2 theta, right?

Answer 12

oh yeah sorry...

Answer 13

now let's just do the top.
\[ \cos 2\theta +i \sin 2 \theta - 1 \]
use the double angle formulas to replace the cos and sin
can you do that and post what you get ?

Answer 14

\[1-2\sin ^{2}\theta + i 2\sin \theta \cos \theta -1\]

Answer 15

ok, I would simplify 1-1, and factor out 2 sin theta

Answer 16

(knowing we want a tan, so try to get sin up top and cos down below)

Answer 17

.

Answer 18

top is \( 2\sin \theta\left( -\sin \theta+ i \cos \theta\right) \)

Answer 19

now do the bottom.
what do you get ?

Answer 20

the same as the top part i fink ,, just we have to factor out cos theta ?

Answer 21

it can't be exactly the same because we have w+1 instead of w-1

Answer 22

yes wait i try it

Answer 23

\[2\cos ^{2}\theta -1 + 2\sin \theta \cos \theta +1\]

Answer 24

good ?

Answer 25

yes, except for the missing i in the 2 sin cos term
so the bottom is
\[ 2 \cos \theta\left( \cos \theta +2 i \sin \theta\right) \]

Answer 26

yes..

Answer 27

**fix that extra 2 I left in the previous post.

you have , so far
\[ \frac{2\sin \theta\left( -\sin \theta+ i \cos \theta\right)}{ 2 \cos \theta\left( \cos \theta + i \sin \theta\right)} \]

Answer 28

yes

Answer 29

there is some obvious things to do: cancel 2/2
write sin/cos as tan

so it is looking closer to what we want
\[ \tan \theta \frac{\left( -\sin \theta+ i \cos \theta\right)}{ \left( \cos \theta + i \sin \theta\right)} \]

any idea what to try next to tackle the fraction ?

Answer 30

no

Answer 31

though I don't know what I'll get up top, I would try multiplying top and bottom by the complex conjugate of the bottom ... because I know that will give me 1 in the bottom)

Answer 32

thank you.. got the answer as per your instructions

Answer 33

There are other ways to do this, but the way we did it is probably as simple as any other.

Answer 34

Since w is defined by \((1,2\theta)\) hence \(x = cos (2\theta), y= sin(2\theta)\)
and \(w = x + iy\\1= 1+oi\\w-1=(x-1)+iy\)
\(w+1= (x+1) +iy\)

\(\dfrac{w-1}{w+1}= \dfrac{(x-1)+iy}{(x+1)+iy}\)

\(=\dfrac{(x-1)(x+1)+y^2}{(x+1)^2+y^2}+i\dfrac{y(x+1)-y(x-1)}{(x+1)^2+y^2}\)
Now, real part:
\(\dfrac{(x-1)(x+1)+y^2}{(x+1)^2+y^2}=\dfrac{x^2-1+y^2}{x^2+2x+y^2+1}\)

but \(x^2+y^2=1\) hence the real part =0
Imaginary part
\(i\dfrac{y(x+1)-y(x-1)}{x^2+y^2+1+2x}=i\dfrac{yx+y-yx+y}{2+2x}\)

\(i\dfrac{2y}{1+x}\)

Answer 35

sorry, the last line is \(=i\dfrac{y}{1+x}\)

Answer 36

Now, replace \(y = sin 2\theta = 2 sin\theta cos\theta\)

\(x = cos 2\theta \\1 + cos 2\theta= 2cos^2 \theta\)

we have it is \(itan\theta\)