anonymous
  • anonymous
someoen helpp pls.. medals and fan question in comment
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
@IrishBoy123
anonymous
  • anonymous
@mathmate

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anonymous
  • anonymous
@phi
phi
  • 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 ?
anonymous
  • anonymous
yes tried it..
anonymous
  • anonymous
i dont know how to use it well.. so can you help in proving ?
phi
  • 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 ?
anonymous
  • anonymous
ive done all that.. and even replaced it in the above eqn
phi
  • phi
what do you get ?
anonymous
  • anonymous
\[(\cos 2\theta +i \sin \theta - 1)/ (\cos 2\theta + isin \theta +1 )\]
phi
  • phi
I assume you mean i sin 2 theta, right?
anonymous
  • anonymous
oh yeah sorry...
phi
  • 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 ?
anonymous
  • anonymous
\[1-2\sin ^{2}\theta + i 2\sin \theta \cos \theta -1\]
phi
  • phi
ok, I would simplify 1-1, and factor out 2 sin theta
phi
  • phi
(knowing we want a tan, so try to get sin up top and cos down below)
IrishBoy123
  • IrishBoy123
.
phi
  • phi
top is \( 2\sin \theta\left( -\sin \theta+ i \cos \theta\right) \)
phi
  • phi
now do the bottom. what do you get ?
anonymous
  • anonymous
the same as the top part i fink ,, just we have to factor out cos theta ?
phi
  • phi
it can't be exactly the same because we have w+1 instead of w-1
anonymous
  • anonymous
yes wait i try it
anonymous
  • anonymous
\[2\cos ^{2}\theta -1 + 2\sin \theta \cos \theta +1\]
anonymous
  • anonymous
good ?
phi
  • phi
yes, except for the missing i in the 2 sin cos term so the bottom is \[ 2 \cos \theta\left( \cos \theta +2 i \sin \theta\right) \]
anonymous
  • anonymous
yes..
phi
  • phi
**fix that extra 2 I left in the previous post. you have , so far \[ \frac{2\sin \theta\left( -\sin \theta+ i \cos \theta\right)}{ 2 \cos \theta\left( \cos \theta + i \sin \theta\right)} \]
anonymous
  • anonymous
yes
phi
  • 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 ?
anonymous
  • anonymous
no
phi
  • 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)
anonymous
  • anonymous
thank you.. got the answer as per your instructions
phi
  • phi
There are other ways to do this, but the way we did it is probably as simple as any other.
Loser66
  • 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 \(i\dfrac{y(x+1)-y(x-1)}{x^2+y^2+1+2x}=i\dfrac{yx+y-yx+y}{2+2x}\) \(i\dfrac{2y}{1+x}\)
Loser66
  • Loser66
sorry, the last line is \(=i\dfrac{y}{1+x}\)
Loser66
  • 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 \(itan\theta\)

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