## anonymous one year ago someoen helpp pls.. medals and fan question in comment

1. anonymous

2. anonymous

@IrishBoy123

3. anonymous

@mathmate

4. anonymous

@phi

5. phi

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 ?

6. anonymous

yes tried it..

7. anonymous

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

8. phi

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 ?

9. anonymous

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

10. phi

what do you get ?

11. anonymous

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

12. phi

I assume you mean i sin 2 theta, right?

13. anonymous

oh yeah sorry...

14. phi

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 ?

15. anonymous

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

16. phi

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

17. phi

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

18. IrishBoy123

.

19. phi

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

20. phi

now do the bottom. what do you get ?

21. anonymous

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

22. phi

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

23. anonymous

yes wait i try it

24. anonymous

$2\cos ^{2}\theta -1 + 2\sin \theta \cos \theta +1$

25. anonymous

good ?

26. phi

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

27. anonymous

yes..

28. phi

**fix that extra 2 I left in the previous post. you have , so far $\frac{2\sin \theta\left( -\sin \theta+ i \cos \theta\right)}{ 2 \cos \theta\left( \cos \theta + i \sin \theta\right)}$

29. anonymous

yes

30. phi

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 ?

31. anonymous

no

32. phi

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)

33. anonymous

34. phi

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

35. Loser66

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 $$i\dfrac{y(x+1)-y(x-1)}{x^2+y^2+1+2x}=i\dfrac{yx+y-yx+y}{2+2x}$$ $$i\dfrac{2y}{1+x}$$

36. Loser66

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

37. Loser66

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 $$itan\theta$$