# Need help on Complex Numbers!

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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
0
6 years ago
#2
is this fp1? and what exam board?
0
6 years ago
#3
Would Euler's identity help?

0
6 years ago
#4
you have to "rationalise" the denominator to get rid of the imaginary component
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#5
(Original post by HarunH1)
is this fp1? and what exam board?
CIE and its Pure Maths 3
0
6 years ago
#6
(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.

0
6 years ago
#7
(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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#8
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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6 years ago
#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?
0
#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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6 years ago
#11
(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?
0
6 years ago
#12
(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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#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 :
(1)(2)(3)
0
6 years ago
#14
(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.
0
6 years ago
#15
(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:
Show
2
6 years ago
#16
There's a useful trick for fractions with terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by (which is 1, obviously), you end up with terms of form which you can easily rewrite in terms of sin and cos theta.
1
6 years ago
#17
(Original post by DFranklin)
There's a useful trick for fractions with terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by (which is 1, obviously), you end up with terms of form 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.
1
6 years ago
#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!
2
#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
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 )
0
#20
(Original post by DFranklin)
There's a useful trick for fractions with terms that can work a fair bit quicker than the "normal" method of rationalising.

Multiply the fraction by (which is 1, obviously), you end up with terms of form 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. 1
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