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# Complex number lock FP2 help! watch

1. Could someone possibly help with 12ii?
I have no idea where to start

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2. (Original post by Mutleybm1996)

Could someone possibly help with 12ii?
I have no idea where to start

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I need the Cartesian equation found earlier first
3. (Original post by Mutleybm1996)
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?
4. (Original post by TeeEm)
I need the Cartesian equation found earlier first

here, thank you

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5. (Original post by Mutleybm1996)

here, thank you

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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
6. (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
Thanks
I tried some bits and bobs but got stuck

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7. (Original post by Mutleybm1996)
Thanks
I tried some bits and bobs but got stuck

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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
8. (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)
9. (Original post by Mutleybm1996)
Changed it, thank you.
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

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.
10. (Original post by TeeEm)
substitute z = x + iy and carry out the algebra

this is straight forward

look at some questions in the link

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 :/

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11. (Original post by Mutleybm1996)
The algebra still isn't working out :/

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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]
12. (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]
If c=a+bi why have you factored(maybe not the right term) it into the real part of the equation?

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13. (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?

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does it say c is complex?
14. (Original post by TeeEm)
does it say c is complex?
"Where C=a+bi For any fixed complex number"

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15. (Original post by Mutleybm1996)
"Where C=a+bi For any fixed complex number"

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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
16. (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
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?

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17. (Original post by Mutleybm1996)
Quick question, why did you substitute those specific values into a and b? Why didn't you switch it round?

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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)
18. (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)
Could you run through what to do after the arg's have been separated? I can't seem to get pi/2

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19. (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

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