a level maths p3 Watch

Aleema Imran
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The complex numbers u and v are given by u = 1 + 2ï3i and v = 3 + 2i. In an Argand diagram,
u and v are represented by the points A and B. A third point C lies in the first quadrant and is
such that BC = 2AB and angle ABC = 90Å. Find the complex number z represented by C, giving
your answer in the form x + iy, where x and y are real and exact

This is a ques from A level Maths Past Paper MJ 17 QP 33. I shall be thankful to anyone who helps me with this one...
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ghostwalker
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(Original post by Aleema Imran)
The complex numbers u and v are given by u = 1 + 2ï3i and v = 3 + 2i. In an Argand diagram,
u and v are represented by the points A and B. A third point C lies in the first quadrant and is
such that BC = 2AB and angle ABC = 90Å. Find the complex number z represented by C, giving
your answer in the form x + iy, where x and y are real and exact

This is a ques from A level Maths Past Paper MJ 17 QP 33. I shall be thankful to anyone who helps me with this one...
Can you link to the orignal question or paper? No idea what MJ 17 means, and you've typos in the question: u = ???
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Aleema Imran
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yeah sure https://papers.gceguide.com/A%20Leve..._s17_qp_33.pdf
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ghostwalker
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(Original post by Aleema Imran)
The complex numbers u and v are given by u = 1 + 2ï3i and v = 3 + 2i. In an Argand diagram,
u and v are represented by the points A and B. A third point C lies in the first quadrant and is
such that BC = 2AB and angle ABC = 90Å. Find the complex number z represented by C, giving
your answer in the form x + iy, where x and y are real and exact

This is a ques from A level Maths Past Paper MJ 17 QP 33. I shall be thankful to anyone who helps me with this one...
I presume you're happy with the diagram.

Since BC is twice AB, I'd start by considering, what's a complex number representing AB?

You now need to rotate that 90 degree anticlockwise to get a complex number in the direction BC. How might you do that in terms of complex numbers?

Then double it to get a complex number representing BC....

And hence C is....
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Aleema Imran
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i can digest the initial part but the rotating part is kinda complex. like all that comes to my mind is that the gradients of two perpendicular lines is equal to -1. can u explain that plz
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ghostwalker
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(Original post by Aleema Imran)
i can digest the initial part but the rotating part is kinda complex. like all that comes to my mind is that the gradients of two perpendicular lines is equal to -1. can u explain that plz
If you have two complex numbers and you multiply them together, then the result arises from multiplying their magnitudes and adding their arguments.

So, in this case what complex number can you multiply by, that leaves the magnitude unchanged and has an argument of 90 degrees (anticlockwise)?
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Aleema Imran
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i understand that but cant get the correct answer. i d be grateful if u could send me a solution? i ve seen the mark scheme and a video on this but there isnt any progress.
(Original post by ghostwalker)
If you have two complex numbers and you multiply them together, then the result arises from multiplying their magnitudes and adding their arguments.

So, in this case what complex number can you multiply by, that leaves the magnitude unchanged and has an argument of 90 degrees (anticlockwise)?
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ghostwalker
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(Original post by Aleema Imran)
i understand that but cant get the correct answer. i d be grateful if u could send me a solution? i ve seen the mark scheme and a video on this but there isnt any progress.
If you understand it, then post your working. And the markscheme - since it's always possible it contains an error.
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Aleema Imran
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i am just making it neater. and i am sending u the link for the mark scheme
https://papers.gceguide.com/A%20Leve..._s17_ms_33.pdf
(Original post by ghostwalker)
If you understand it, then post your working. And the markscheme - since it's always possible it contains an error.
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Aleema Imran
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heres the solution. i hope u now get a fair idea of the point i am stuck at
(Original post by Aleema Imran)
i am just making it neater. and i am sending u the link for the mark scheme
https://papers.gceguide.com/A%20Leve..._s17_ms_33.pdf
Attached files
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ghostwalker
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(Original post by Aleema Imran)
heres the solution. i hope u now get a fair idea of the point i am stuck at
OK, your w corresponds to BA, rather than AB - which is fine.

And you've scaled it by 2, to get the desired length, in -4+(4\sqrt{3}-4)i

Since you've worked out w using BA, you need to rotate this 90 degrees clockwise, to get BC. And you don't wish to change it's length further, so we multiply our 2w by -i (the complex number with magnitude 1, and arg minus 90 degrees)

This gives us BC.

Then we simply add on the complex number corresponding to B, to get the complex number corresponding to C, and we're done.
Last edited by ghostwalker; 4 days ago
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Aleema Imran
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oh okay thank u so much. i finally obtained the correct ans thank God. but can u explain one thing to me? y r we multiplying 2w by -1

(Original post by ghostwalker)
OK, your w corresponds to BA, rather than AB - which is fine.

And you've scaled it by 2, to get the desired length, in -4+(4\sqrt{3}-4)i

Since you've worked out w using BA, you need to rotate this 90 degrees clockwise, to get BC. And you don't wish to change it's length further, so we multiply our 2w by -i (the complex number with magnitude 1, and arg minus 90 degrees)

This gives us BC.

Then we simply add on the complex number corresponding to B, to get the complex number corresponding to C, and we're done.
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ghostwalker
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(Original post by Aleema Imran)
oh okay thank u so much. i finally obtained the correct ans thank God. but can u explain one thing to me? y r we multiplying 2w by -1
It's -i, not -1

If you consider a complex number in magnitude argument form. E.g let x=a\angle b and y=c\angle d

Then the product of the two numbers xy=ac \angle (b+d). Their magnitudes are multiplied and their angles are added.

In this case the complex number w, corresponds to a vector BA - it has magnitude and direction.

We need to rotate it 90 degrees clockwise. So, we multiply by 1\angle -90, which is just "-i", the complex number with magnitude 1, and 90 degrees clockwise to the +ve x axis.
Last edited by ghostwalker; 4 days ago
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Aleema Imran
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oh so its actually like z = r (cos theeta + i sin theeta) where r is its magnitude and theeta is its arg. cos -90 is zero and sin -90 is -1 and -1 multiplied with i is -1. magnitude is the same so r = 1? i think it makes sense now???
(Original post by ghostwalker)
It's -i, not -1

If you consider a complex number in magnitude argument form. E.g let x=a\angle b and y=c\angle d

Then the product of the two numbers xy=ac \angle (b+d). Their magnitudes are multiplied and their angles are added.

In this case the complex number w, corresponds to a vector BA - it has magnitude and direction.

We need to rotate it 90 degrees clockwise. So, we multiply by 1\angle -90, which is just "-i", the complex number with magnitude 1, and 90 degrees clockwise to the +ve x axis.
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ghostwalker
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(Original post by Aleema Imran)
oh so its actually like z = r (cos theeta + i sin theeta) where r is its magnitude and theeta is its arg. cos -90 is zero and sin -90 is -1 and -1 multiplied with i is -1. magnitude is the same so r = 1? i think it makes sense now???
In red. -1\times i = -i
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Aleema Imran
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oh thank u so much. i finally get it.
(Original post by ghostwalker)
In red. -1\times i = -i
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Aleema Imran
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ummmm.....i need some help with physics
https://papers.gceguide.com/A%20Leve..._m18_qp_22.pdf
i cant solve part b (v) (the change in momentum part) of ques # 2. somebody help plz? thanks in advancesomeb
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