Need help on Complex Numbers!

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annie aw
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#1
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#1
actual qs. : The complex number w has modulus 1 and argument 2θ radians. show that (w-1)/(w+1) = itanθ


simplify
( cos 2θ + i(sin 2θ) - 1) ÷ ( cos 2θ + i(sin 2θ) + 1 )

show that it is equal to = i tanθ
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HarunH1
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#2
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is this fp1? and what exam board?
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ChaoticButterfly
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#3
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#3
Would Euler's identity help?

Image
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tanyapotter
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#4
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you have to "rationalise" the denominator to get rid of the imaginary component
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annie aw
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#5
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#5
(Original post by HarunH1)
is this fp1? and what exam board?
CIE and its Pure Maths 3
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ChaoticButterfly
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#6
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(Original post by annie aw)
yes, but I have applied it to the equation. (edited on my qs)
dunno I left all this **** behind when I graduated.

on your own kiddo
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username1560589
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(Original post by annie aw)
actual qs. : The complex number w has modulus 1 and argument 2θ radians. show that (w-1)/(w+1) = itanθ


simplify
( cos 2θ + i(sin 2θ) - 1) ÷ ( cos 2θ + i(sin 2θ) + 1 )

show that it is equal to = i tanθ
I did it by converting into polar form and rationalising the denominator. How far have you gotten?

(Original post by _Bembridge)
...
No. A different kind of rationalising and you made a mistake.
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annie aw
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#8
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i got till the part expanding the double angles.. then i got confused

(Original post by morgan8002)
I did it by converting into polar form and rationalising the denominator. How far have you gotten?



No. A different kind of rationalising and you made a mistake.
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username1560589
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#9
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#9
(Original post by annie aw)
i got till the part expanding the double angles.. then i got confused
Can you type up your last line?
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annie aw
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#10
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#10
( 1 - 2sin^2θ + i (2sinθcosθ) + 1 ) ÷ ( 2cos^2θ - 1 + i (2sinθcosθ) + 1)

(Original post by morgan8002)
Can you type up your last line?
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username1560589
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#11
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(Original post by annie aw)
( 1 - 2sin^2θ + i (2sinθcosθ) + 1 ) ÷ ( 2cos^2θ - 1 + i (2sinθcosθ) + 1)
I don't get that. What did you multiply by to rationalise the denominator?
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Zacken
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#12
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(Original post by morgan8002)
I don't get that. What did you multiply by to rationalise the denominator?
She hasn't multiplied anything, simply converted double angles into single angles for some obscure reason.
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annie aw
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#13
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#13
(Original post by morgan8002)
I don't get that. What did you multiply by to rationalise the denominator?
i used the double angle formula :
ImageImageImage(1)ImageImageImage(2)ImageImageImage(3)ImageImageImage
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username1560589
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#14
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(Original post by annie aw)
( 1 - 2sin^2θ + i (2sinθcosθ) + 1 ) ÷ ( 2cos^2θ - 1 + i (2sinθcosθ) + 1)
(Original post by annie aw)
i used the double angle formula :
ImageImageImage(1)ImageImageImage(2)ImageImageImage(3)ImageImageImage
The + sign bolded above should be a - sign.
Rationalise the denominator, then use double angle fomulae if necessary. Otherwise it will get overcomplicated.
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Dingooose
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#15
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(Original post by annie aw)
actual qs. : The complex number w has modulus 1 and argument 2θ radians. show that (w-1)/(w+1) = itanθ


simplify
( cos 2θ + i(sin 2θ) - 1) ÷ ( cos 2θ + i(sin 2θ) + 1 )

show that it is equal to = i tanθ
There are probably more effective solutions but here is mine (refer to it only once you're done):
Spoiler:
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DFranklin
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#16
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#16
There's a useful trick for fractions with (1 \pm e^{2i \theta}) terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by \dfrac{e^{-i\theta}}{e^{-i\theta}} (which is 1, obviously), you end up with terms of form e^{i\theta} \pm e^{-i\theta} which you can easily rewrite in terms of sin and cos theta.
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Dingooose
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#17
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(Original post by DFranklin)
There's a useful trick for fractions with (1 \pm e^{2i \theta}) terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by \dfrac{e^{-i\theta}}{e^{-i\theta}} (which is 1, obviously), you end up with terms of form e^{i\theta} \pm e^{-i\theta} which you can easily rewrite in terms of sin and cos theta.
You posted that one minute after I posted my method... Now I feel silly for complicating it.
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DFranklin
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#18
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#18
(Original post by Dingooose)
You posted that one minute after I posted my method... Now I feel silly for complicating it.
If it makes you feel better I did it your way for about 10 years before someone posted this trick!
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annie aw
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#19
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#19
(Original post by Dingooose)
There are probably more effective solutions but here is mine (refer to it only once you're done):
Spoiler:
Show
Name:  Capture.PNG
Views: 897
Size:  6.8 KB
Thanks! I used the same method as yours, and was stuck at simplifying it.. (sorry i actually took a peek before i'm actually done)
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annie aw
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#20
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#20
(Original post by DFranklin)
There's a useful trick for fractions with (1 \pm e^{2i \theta}) terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by \dfrac{e^{-i\theta}}{e^{-i\theta}} (which is 1, obviously), you end up with terms of form e^{i\theta} \pm e^{-i\theta} which you can easily rewrite in terms of sin and cos theta.
I tried your method, right after I finished the first try. And it really saves all the troubles! thank you.
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