# a level maths p3

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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...

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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#2

(Original post by

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...

**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...

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Aleema Imran

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ghostwalker

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**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...

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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#5

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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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

**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

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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#7

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

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)?

**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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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.

**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.

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#9

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

https://papers.gceguide.com/A%20Leve..._s17_ms_33.pdf

(Original post by

If you understand it, then post your working. And the markscheme - since it's always possible it contains an error.

**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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#10

heres the solution. i hope u now get a fair idea of the point i am stuck at

(Original post by

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

**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

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#11

(Original post by

heres the solution. i hope u now get a fair idea of the point i am stuck at

**Aleema Imran**)heres the solution. i hope u now get a fair idea of the point i am stuck at

And you've scaled it by 2, to get the desired length, in

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; 3 years ago

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#12

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

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

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.

**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

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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(Original post by

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

**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

If you consider a complex number in magnitude argument form. E.g let and

Then the product of the two numbers . 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 , which is just "-i", the complex number with magnitude 1, and 90 degrees clockwise to the +ve x axis.

Last edited by ghostwalker; 3 years ago

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#14

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

It's -i, not -1

If you consider a complex number in magnitude argument form. E.g let and

Then the product of the two numbers . 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 , which is just "-i", the complex number with magnitude 1, and 90 degrees clockwise to the +ve x axis.

**ghostwalker**)It's -i, not -1

If you consider a complex number in magnitude argument form. E.g let and

Then the product of the two numbers . 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 , which is just "-i", the complex number with magnitude 1, and 90 degrees clockwise to the +ve x axis.

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(Original post by

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???

**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???

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#16

oh thank u so much. i finally get it.

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#17

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

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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