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# Further complex numbers (transformations) Watch

1. Hi all, I'm doing this question:

I can do part a and b fine, but not c.

I made z the subject, then plugged in z=x+iy and w =u+iv then rationalised the denominator and equated the imaginary part to 0. but it doesn't seem to give me the right circle equation. I end up with a weird circle equation in terms of u and v.

Am I working backwards or something?
2. (Original post by Sir_Malc)
Hi all, I'm doing this question:

I can do part a and b fine, but not c.

I made z the subject, then plugged in z=x+iy and w =u+iv then rationalised the denominator and equated the imaginary part to 0. but it doesn't seem to give me the right circle equation. I end up with a weird circle equation in terms of u and v.

Am I working backwards or something?
There was no need to introduce , just . This should clear it all up!
3. (Original post by Sir_Malc)
Hi all, I'm doing this question:

I can do part a and b fine, but not c.

I made z the subject, then plugged in z=x+iy and w =u+iv then rationalised the denominator and equated the imaginary part to 0. but it doesn't seem to give me the right circle equation. I end up with a weird circle equation in terms of u and v.

Am I working backwards or something?
Messy way: set and find out what the imaginary part of is in terms of a and b. Now do the same for .
4. (Original post by Zacken)
There was no need to introduce , just . This should clear it all up!
How come you only need the z=x+iy?

In the Edexcel book there is similar examples and they always seem to rearrange the transformation first to make z the subject. then plug in z and w.
5. (Original post by Gregorius)
Messy way: set and find out what the imaginary part of is in terms of a and b. Now do the same for .
Not so messy because you're already asked to find the cartesian equation in part (a).

Out of interest, is there a nice way to do stuff like this? I've glanced over a few complex analysis notes and they always talk about "circles going to lines" and "lines going to lines" or "mobious transforms" or such and I thought it might be vaguely related to this sort of thing where circles in the -plane go to lines in the -plane, etc...
6. (Original post by Sir_Malc)
How come you only need the z=x+iy?

In the Edexcel book there is similar examples and they always seem to rearrange the transformation first to make z the subject. then plug in z and w.
Well, I suppose you could plug in but it would serve no purpose since you'll end up only considering anyway. The video I linked works through this precise example, if you're interested.
7. (Original post by Zacken)
Not so messy because you're already asked to find the cartesian equation in part (a).

Out of interest, is there a nice way to do stuff like this? I've glanced over a few complex analysis notes and they always talk about "circles going to lines" and "lines going to lines" or "mobious transforms" or such and I thought it might be vaguely related to this sort of thing where circles in the -plane go to lines in the -plane, etc...
Yes, mobius transformations do have very nice properties, one of which is to map (generalized) circles to (generalized) circles - where a generalized circle is either a common or garden circle or a circle of infinite radius (i.e. a straight line). Once you know this, the question can be dispatched into the long grass by inspecting the fate of three points.
8. (Original post by Gregorius)
Yes, mobius transformations do have very nice properties, one of which is to map (generalized) circles to (generalized) circles - where a generalized circle is either a common or garden circle or a circle of infinite radius (i.e. a straight line). Once you know this, the question can be dispatched into the long grass by inspecting the fate of three points.
Ahhh. Yes, I think I've heard about that - you usually try and pick 'easy points' like 0 or infinity or 1 (where infinity is the thingy at the top of the Riemann sphere thingymabob) - it's all very interesting, cheers!
9. (Original post by Zacken)
Well, I suppose you could plug in but it would serve no purpose since you'll end up only considering anyway. The video I linked works through this precise example, if you're interested.
So I should get the same answer even if I rearrange the transformation to z = (i-3w)/(wi-1)

then plug in z=x+iy and w=u+iv

and then equate the imaginary part to 0. Then I should get a similar equation but in terms of u and v right?
10. (Original post by Sir_Malc)
So I should get the same answer even if I rearrange the transformation to z = (i-3w)/(wi-1)

then plug in z=x+iy and w=u+iv

and then equate the imaginary part to 0. Then I should get a similar equation but in terms of u and v right?
You'll get it in terms of u and v, but that's going to serve of no use to you because you want to prove a fact about the (x-y) plane and not the (u-v) plane.
11. (Original post by Zacken)
You'll get it in terms of u and v, but that's going to serve of no use to you because you want to prove a fact about the (x-y) plane and not the (u-v) plane.
Thankyou. This makes sense to me now/
12. (Original post by Sir_Malc)
Thankyou. This makes sense to me now/
Cheers!

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