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

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phi
 one year ago
Best ResponseYou've already chosen the best response.2my 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 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont know how to use it well.. so can you help in proving ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2First, do the top w1 replace w with exp(i 2 theta) and then use Euler to rewrite in terms of cos and sin can you do that ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ive done all that.. and even replaced it in the above eqn

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\cos 2\theta +i \sin \theta  1)/ (\cos 2\theta + isin \theta +1 )\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2I assume you mean i sin 2 theta, right?

phi
 one year ago
Best ResponseYou've already chosen the best response.2now 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 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[12\sin ^{2}\theta + i 2\sin \theta \cos \theta 1\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2ok, I would simplify 11, and factor out 2 sin theta

phi
 one year ago
Best ResponseYou've already chosen the best response.2(knowing we want a tan, so try to get sin up top and cos down below)

phi
 one year ago
Best ResponseYou've already chosen the best response.2top is \( 2\sin \theta\left( \sin \theta+ i \cos \theta\right) \)

phi
 one year ago
Best ResponseYou've already chosen the best response.2now do the bottom. what do you get ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the same as the top part i fink ,, just we have to factor out cos theta ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2it can't be exactly the same because we have w+1 instead of w1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2\cos ^{2}\theta 1 + 2\sin \theta \cos \theta +1\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2yes, except for the missing i in the 2 sin cos term so the bottom is \[ 2 \cos \theta\left( \cos \theta +2 i \sin \theta\right) \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2**fix that extra 2 I left in the previous post. you have , so far \[ \frac{2\sin \theta\left( \sin \theta+ i \cos \theta\right)}{ 2 \cos \theta\left( \cos \theta + i \sin \theta\right)} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2there 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 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2though 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you.. got the answer as per your instructions

phi
 one year ago
Best ResponseYou've already chosen the best response.2There are other ways to do this, but the way we did it is probably as simple as any other.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Since w is defined by \((1,2\theta)\) hence \(x = cos (2\theta), y= sin(2\theta)\) and \(w = x + iy\\1= 1+oi\\w1=(x1)+iy\) \(w+1= (x+1) +iy\) \(\dfrac{w1}{w+1}= \dfrac{(x1)+iy}{(x+1)+iy}\) \(=\dfrac{(x1)(x+1)+y^2}{(x+1)^2+y^2}+i\dfrac{y(x+1)y(x1)}{(x+1)^2+y^2}\) Now, real part: \(\dfrac{(x1)(x+1)+y^2}{(x+1)^2+y^2}=\dfrac{x^21+y^2}{x^2+2x+y^2+1}\) but \(x^2+y^2=1\) hence the real part =0 Imaginary part \(i\dfrac{y(x+1)y(x1)}{x^2+y^2+1+2x}=i\dfrac{yx+yyx+y}{2+2x}\) \(i\dfrac{2y}{1+x}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0sorry, the last line is \(=i\dfrac{y}{1+x}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Now, 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 \(itan\theta\)
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