Complex number lock FP2 help!
Could someone possibly help with 12ii?
I have no idea where to start
Posted from TSR Mobile
Scroll to see replies
Original post by Mutleybm1996
I need the Cartesian equation found earlier first
Original post by Mutleybm1996
Could someone possibly help with 12ii?
I have no idea where to start
Could someone possibly help with 12ii?
I have no idea where to start
1. By the way arg works we have:
2. is the complex number that takes us from fixed point a to arbitrary point z. Here z is restricted to the circumference of a circle.
3. measures the angle that this complex number makes with the +ve x-axis.
4. By circle geometry, you should know that the angle subtended by any point on the circumference of a circle between two fixed points (i.e. in an arc) is constant.
Does that give you a clue?
Original post by TeeEm
I need the Cartesian equation found earlier first
here, thank you
Posted from TSR Mobile
(edited 9 years ago)
Original post by Mutleybm1996
arg[(z-a)/(z-b)]=θ
arg(z-a)- arg(z-b)=θ
arg(z-6-√3) -arg(z-6+√3) = θ
sub the point they give you in the above equation to get θ
(edited 9 years ago)
Original post by TeeEm
arg[(z-a)/(z-b)]=θ
arg(z-a)- arg(z-b)=θ
arg(z-6-3i) -arg(z-6+3i) = θ
sub the point they give you in the above equation to get θ
arg(z-a)- arg(z-b)=θ
arg(z-6-3i) -arg(z-6+3i) = θ
sub the point they give you in the above equation to get θ
Thanks
Could i also ask you about q9?
I tried some bits and bobs but got stuck
Posted from TSR Mobile
Original post by Mutleybm1996
my reply is incorrect as I am skipping between pages look at the correction below
arg[(z-a)/(z-b)]=θ
arg(z-a)- arg(z-b)=θ
arg(z-6-3√3) -arg(z-6+3√3) = θ
sub the point they give you in the above equation to get θ
you should get theta = pi/2
(edited 9 years ago)
Original post by TeeEm
my reply is incorrect as I am skipping between pages look at the correction below
Changed it, thank you.
Could you possibly look at Q9? (In a post earlier)
Original post by Mutleybm1996
Changed it, thank you.
Could you possibly look at Q9? (In a post earlier)
Could you possibly look at Q9? (In a post earlier)
substitute z = x + iy and carry out the algebra
this is straight forward
look at some questions in the link
http://www.madasmaths.com/archive/maths_booklets/further_topics/various/complex_numbers_part_2_exam_questions.pdf
there many questions with solutions oncomplex but they are mixed in topics.
At least 40 questions on complex loci all mixed up/the booklet gets progressively harder.
Original post by TeeEm
substitute z = x + iy and carry out the algebra
this is straight forward
look at some questions in the link
http://www.madasmaths.com/archive/maths_booklets/further_topics/various/complex_numbers_part_2_exam_questions.pdf
there many questions with solutions oncomplex but they are mixed in topics.
At least 40 questions on complex loci all mixed up/the booklet gets progressively harder.
this is straight forward
look at some questions in the link
http://www.madasmaths.com/archive/maths_booklets/further_topics/various/complex_numbers_part_2_exam_questions.pdf
there many questions with solutions oncomplex but they are mixed in topics.
At least 40 questions on complex loci all mixed up/the booklet gets progressively harder.
The algebra still isn't working out :/
Posted from TSR Mobile
Original post by Mutleybm1996
here are the first three lines
then you should be able to complete it
|x+iy-k2c|=k|x+iy-c|
|(x-k2c)+iy|=k|(x-c)+iy|
(x-k2c)2+y2=k2[(x-c)2+y2]
Original post by TeeEm
here are the first three lines
then you should be able to complete it
|x+iy-k2c|=k|x+iy-c|
|(x-k2c)+iy|=k|(x-c)+iy|
(x-k2c)2+y2=k2[(x-c)2+y2]
then you should be able to complete it
|x+iy-k2c|=k|x+iy-c|
|(x-k2c)+iy|=k|(x-c)+iy|
(x-k2c)2+y2=k2[(x-c)2+y2]
If c=a+bi why have you factored(maybe not the right term) it into the real part of the equation?
Posted from TSR Mobile
Original post by Mutleybm1996
If c=a+bi why have you factored(maybe not the right term) it into the real part of the equation?
Posted from TSR Mobile
Posted from TSR Mobile
does it say c is complex?
Original post by TeeEm
does it say c is complex?
"Where C=a+bi For any fixed complex number"
Posted from TSR Mobile
Original post by Mutleybm1996
sorry I did not see it
the method is still the same
put z=x+iy
c = a +bi
group real and imaginary inside the mods
"apply" the mods
square both sides (careful you do not forget to square the k on the right and then multiply the whole RHS by k2)
then a big mess to simplify.
this is nothing but an exercise in concentration.
good luck
Original post by TeeEm
arg[(z-a)/(z-b)]=θ
arg(z-a)- arg(z-b)=θ
arg(z-6-√3) -arg(z-6+√3) = θ
sub the point they give you in the above equation to get θ
arg(z-a)- arg(z-b)=θ
arg(z-6-√3) -arg(z-6+√3) = θ
sub the point they give you in the above equation to get θ
Quick question, why did you substitute those specific values into a and b? Why didn't you switch it round?
EDIT: sorted that part
Okay, so I've split it into the two arg sections, what do I do now?
Posted from TSR Mobile
(edited 9 years ago)
Original post by Mutleybm1996
Quick question, why did you substitute those specific values into a and b? Why didn't you switch it round?
Posted from TSR Mobile
Posted from TSR Mobile
answer to (a) this is where the arc meets the x axis
answer to (b) if I put it these values the other way round theta will be -pi/2 which gives exactly the same equation, if you multiply it by -1 (I am not sure if this is what you are asking)
Original post by TeeEm
answer to (a) this is where the arc meets the x axis
answer to (b) if I put it these values the other way round theta will be -pi/2 which gives exactly the same equation, if you multiply it by -1 (I am not sure if this is what you are asking)
answer to (b) if I put it these values the other way round theta will be -pi/2 which gives exactly the same equation, if you multiply it by -1 (I am not sure if this is what you are asking)
Could you run through what to do after the arg's have been separated? I can't seem to get pi/2
Posted from TSR Mobile
Original post by Mutleybm1996
Could you run through what to do after the arg's have been separated? I can't seem to get pi/2
Posted from TSR Mobile
Posted from TSR Mobile
wait 2 minutes please
Quick Reply
Related discussions
- Lock & Key!
- Further Maths 2023/2024!!!
- Further maths - FP1, FP2, FM1 second year, after FM1, FS1 1st year
- Complex numbers A level question
- I’m so confused about part B Argand diagrams
- Cyberstart Intern Level 3 Challenge 2
- AQA A-level computer science paper 1
- Integral Maths - Complex numbers - Test C1
- Road to a 2:1 - 2nd year maths GYG
- A-level Philosophy Study Group 2022-2023
- A Level furthermaths
- a level math
- Engineering complex numbers + trig help
- inequality involving modulus
- PwC Game Based Assessment 2023 intake Graduate Scheme
- Messed up CIE Math P3 and Physics P4.
- Medical schools application advice
- An inequation
- Lost my keys in uni halls
- Accommodation advice
Latest
Last reply 47 minutes ago
Official University College London Applicant Thread for 2024Last reply 52 minutes ago
Can I do economics degree without a level maths?Last reply 1 hour ago
How to choose unis in UCAS application for CS undergraduate course?Last reply 1 hour ago
Rishi Sunak pledges to remove benefits for people not taking jobs after 12 monthsLast reply 1 hour ago
Got my crush's number 2 days ago and no reply yet. What could be the reasons?Posted 2 hours ago
GB News set to axe 40 jobs after channel posts heavy lossesLast reply 2 hours ago
Rwanda bill likely to be stalled at least till April after seven defeats in the LordsPosted 2 hours ago
Sunak rejects offer of youth mobility scheme between EU and UKTrending
Last reply 1 week ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
Trending
Last reply 1 week ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
