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Further Maths: Modulus + Argument

If r = cis(theta), find the real and imaginary parts of (r-1)/(r+1) in their simplest forms

I got that Re(r) = 0, but have no clue for the imaginary part.
Original post by Harry-Pikesley
If r = cis(theta), find the real and imaginary parts of (r-1)/(r+1) in their simplest forms

I got that Re(r) = 0, but have no clue for the imaginary part.


You need to convert r−1r+1\dfrac{r-1}{r+1} into the form A+iBA+iB first.

So rewrite the fraction in terms of cos⁡θ\cos \theta and sin⁡θ\sin \theta and proceed.

P.S. Re(r) = 0 doesn't make sense though... since r=cos⁡θ+isin⁡θr = \cos \theta + i \sin \theta so obviously ℜ(r)=cos⁡θ\Re (r) = \cos \theta.
(edited 5 years ago)
Original post by RDKGames
You need to convert r−1r+1\dfrac{r-1}{r+1} into the form A+iBA+iB first.

So rewrite the fraction in terms of cos⁡θ\cos \theta and sin⁡θ\sin \theta and proceed.

P.S. Re(r) = 0 doesn't make sense though... since r=cos⁡θ+isin⁡θr = \cos \theta + i \sin \theta so obviously ℜ(r)=cos⁡θ\Re (r) = \cos \theta.


I meant to say that the real part of the fraction is equal to 0, I believe.
Original post by Harry-Pikesley
I meant to say that the real part of the fraction is equal to 0, I believe.


Alright, post your working out and we can find out.
Original post by Harry-Pikesley
I meant to say that the real part of the fraction is equal to 0, I believe.


(r-1)/(r+1) = (cos + isin -1)/(cos + isin + 1)

Then what?
Original post by Harry-Pikesley
(r-1)/(r+1) = (cos + isin -1)/(cos + isin + 1)

Then what?


You don't want a complex denominator... so do something to that fraction that is analogous to 'rationalising the denominator'
Original post by RDKGames
You don't want a complex denominator... so do something to that fraction that is analogous to 'rationalising the denominator'

I get that I need to rationalise, but which term from the denominator changes sign when it is in the fraction used to rationalise? The minus one to plus one?
Original post by Harry-Pikesley
I get that I need to rationalise, but which term from the denominator changes sign when it is in the fraction used to rationalise? The minus one to plus one?


No, you change ii to −i-i since that's what you aim to eliminate.
Original post by RDKGames
No, you change ii to −i-i since that's what you aim to eliminate.

I simplified and got (2isintheta)/(costheta + 2) but thats probably wrong
Original post by Harry-Pikesley
I simplified and got (2isintheta)/(costheta + 2) but thats probably wrong


Yep, that's wrong. Check the coefficient of cos⁡θ\cos \theta.
Original post by RDKGames
Yep, that's wrong. Check the coefficient of cos⁡θ\cos \theta.

Coefficient is 2?
Original post by Harry-Pikesley
Coefficient is 2?


Yes
Original post by RDKGames
Yes

Cool, I've solved it now. Thanks

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