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Need help on Complex Numbers!

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
(edited 8 years ago)

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Reply 1
Avatar for HarunH1
HarunH1
13
is this fp1? and what exam board?
Would Euler's identity help?

(edited 8 years ago)
you have to "rationalise" the denominator to get rid of the imaginary component
Reply 4
Avatar for annie aw
annie aw
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1
Original post by HarunH1
is this fp1? and what exam board?


CIE and its Pure Maths 3
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
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.
Reply 7
Avatar for annie aw
annie aw
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1
i got till the part expanding the double angles.. then i got confused :s-smilie:

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.
Original post by annie aw
i got till the part expanding the double angles.. then i got confused :s-smilie:


Can you type up your last line?
Reply 9
Avatar for annie aw
annie aw
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1
( 1 - 2sin^2θ + i (2sinθcosθ) + 1 ) ÷ ( 2cos^2θ - 1 + i (2sinθcosθ) + 1)

Original post by morgan8002
Can you type up your last line?
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?
Reply 11
Avatar for Zacken
Zacken
22
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.
Reply 12
Avatar for annie aw
annie aw
OP
1
Original post by morgan8002
I don't get that. What did you multiply by to rationalise the denominator?


i used the double angle formula :
(1)(2)(3)
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 :
(1)(2)(3)


The + sign bolded above should be a - sign.
Rationalise the denominator, then use double angle fomulae if necessary. Otherwise it will get overcomplicated.
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

There's a useful trick for fractions with (1±e2iθ)(1 \pm e^{2i \theta}) terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by e−iθe−iθ\dfrac{e^{-i\theta}}{e^{-i\theta}} (which is 1, obviously), you end up with terms of form eiθ±e−iθe^{i\theta} \pm e^{-i\theta} which you can easily rewrite in terms of sin and cos theta.
Original post by DFranklin
There's a useful trick for fractions with (1±e2iθ)(1 \pm e^{2i \theta}) terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by e−iθe−iθ\dfrac{e^{-i\theta}}{e^{-i\theta}} (which is 1, obviously), you end up with terms of form eiθ±e−iθe^{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.
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!
Reply 18
Avatar for annie aw
annie aw
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Original post by Dingooose
There are probably more effective solutions but here is mine (refer to it only once you're done):

Spoiler



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:colondollar:)
Reply 19
Avatar for annie aw
annie aw
OP
1
Original post by DFranklin
There's a useful trick for fractions with (1±e2iθ)(1 \pm e^{2i \theta}) terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by e−iθe−iθ\dfrac{e^{-i\theta}}{e^{-i\theta}} (which is 1, obviously), you end up with terms of form eiθ±e−iθe^{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. :biggrin:

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