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Edexcel FP2 complex transformations

Parts a and b are simple enough, but I don't understand how to do part c. I've taken a look at the mark scheme, but it's incredibly vague and skips steps.
Anyone care to explain this to me?
http://oi62.tinypic.com/oavy1y.jpg
(edited 8 years ago)
Reply 1
I just spent quite a while trying to do that through real and imaginary parts but I've done it now and hopefully can be of assistance :colondollar:

Basically you need to manipulate the algebra to get (z - 2i)/(z + 2) in terms of z. This is best done by first rearranging to find z. After some fractional manipulation to reach that you should have it equal to a very nicely simplified w function, whose argument naturally must be pi/2. Or is your problem with doing said algebraic working/interpreting the final function to sketch it?
Reply 2
Original post by 1 8 13 20 42
I just spent quite a while trying to do that through real and imaginary parts but I've done it now and hopefully can be of assistance :colondollar:

Basically you need to manipulate the algebra to get (z - 2i)/(z + 2) in terms of z. This is best done by first rearranging to find z. After some fractional manipulation to reach that you should have it equal to a very nicely simplified w function, whose argument naturally must be pi/2. Or is your problem with doing said algebraic working/interpreting the final function to sketch it?


I could see that was what the end goal was, but didn't see how they got there. I understand where the (z-2i) comes from, but not the (z+2) after the rearranging. I'm just having one of those days...
Reply 3
Original post by H0PEL3SS
I could see that was what the end goal was, but didn't see how they got there. I understand where the (z-2i) comes from, but not the (z+2) after the rearranging. I'm just having one of those days...


post the question please
Reply 4
Original post by TeeEm
post the question please

Had a weird problem with putting it directly into a post, the link is in the OP.
Reply 5
Original post by H0PEL3SS
Had a weird problem with putting it directly into a post, the link is in the OP.


:smile:Sorry but I am scared of viruses ...
besides the whole point of being here is not to go to a third party site if we can help it.
Reply 6


Original post by TeeEm
:smile:Sorry but I am scared of viruses ...besides the whole point of being here is not to go to a third party site if we can help it.



Well, it worked now, thankfully :smile:
Reply 7
Original post by H0PEL3SS





Well, it worked now, thankfully :smile:


which part?
Reply 8
Original post by TeeEm
which part?


Part c. I can do the other two, but don't quite understand how they've rearranged the the transformation for (z-2i)/(z+2) from that alone.
Reply 9
Original post by H0PEL3SS
Part c. I can do the other two, but don't quite understand how they've rearranged the the transformation for (z-2i)/(z+2) from that alone.



it can be done in two ways that I can think of

Try this
solve the transformation for z
subtract -1 + i (centre of arc in previous part)
mod both sides
LHS =root2 (radius of the arc in previous part)
RHS set w = u+ iv
....finish off
(edited 8 years ago)
Reply 10
Original post by TeeEm
it can be done in two ways that I can think of

Try this
solve the transformation for z
subtract 1 + i (centre of arc in previous part)
mod both sides
LHS =root2 (radius of the arc in previous part)
RHS set w = u+ iv
....finish off

Thank you. I'll have an attempt tomorrow. The mark scheme was incredibly sparse and didn't mention any alternative methods in detail.
:smile:
Reply 11
Original post by H0PEL3SS
Thank you. I'll have an attempt tomorrow. The mark scheme was incredibly sparse and didn't mention any alternative methods in detail.
:smile:


see my correction centre is -1 + i
Original post by H0PEL3SS
Part c. I can do the other two, but don't quite understand how they've rearranged the the transformation for (z-2i)/(z+2) from that alone.


A (half) line in the w-plane will be of the form arg(wa)=c\arg(w-a)=c where aa is complex and cc is an angle. But you already have an expression of the form arg(something in the z-plane)=c\arg (\text{something in the z-plane}) = c. This should give you a hint as to what to do.

[edit: The numerically named poster suggested this earlier I think, now i read it properly.]
(edited 8 years ago)

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