# A level Further Maths Complex number Help

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#1
In an Argand diagram, the points A, B and C are the vertices of an equilateral triangle with its centre at the origin. The point A represents the complex number 6 + 2i.
(a) Find the complex numbers represented by the points B and C, giving your answers in the form x + iy, where x and y are real and exact.

The points D, E and F are the midpoints of the sides of triangle ABC.
(b) Find the exact area of triangle DEF.

I don't really know if I got the first part right since I can't get a nice exact form for it. I got 2root10 (cos(arctan(1/3)+2pile/3) + isin(arctan(1/3)+2pile/3) and 2root10 (cos(arctan(1/3)+4pile/3) + isin(arctan(1/3)+4pile/3).
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2 weeks ago
#2
You could consider each point as a rotation which gets rid of the angle sum inside the trig function.
Last edited by mqb2766; 2 weeks ago
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#3
(Original post by mqb2766)
You could consider each point as a rotation which gets rid of the angle sum inside the trig function.
Sorry but I don't quite understand what you mean Like how can I turn those point into rotations???
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2 weeks ago
#4
(Original post by cm49919)
Sorry but I don't quite understand what you mean Like how can I turn those point into rotations???
A rotation by phi is multiplication by
e^(i*phi)

So either
* Create a "normal" equilateral triangle where the vertices are sqrt(40) from the origin and one vertex lies in the +ive x axis. Then rotate each vertex by atan(1/3).
* Rotate 6+2i by phi=2pi/3 (twice)

As they want an exact answer in rectangular form, maybe work directly in that, and Id try the second approach first as it seems slightly easier, but neither should be hard.
Last edited by mqb2766; 2 weeks ago
1
#5
(Original post by mqb2766)
A rotation by phi is multiplication by
e^(i*phi)

So either
* Create a "normal" equilateral triangle where the vertices are sqrt(40) from the origin and one vertex lies in the +ive x axis. Then rotate each vertex by atan(1/3).
* Rotate 6+2i by phi=2pi/3 (twice)

As they want an exact answer in rectangular form, maybe work directly in that, and Id try the second approach first as it seems slightly easier, but neither should be hard.
I have tried the second method by multiply 6+2i with e^(i*2pi/3) and I just got the answer same as the answer I mentioned at the start since we have to also turn 6+2i into the form re^(i*phi), isn't it????
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2 weeks ago
#6
(Original post by cm49919)
I have tried the second method by multiply 6+2i with e^(i*2pi/3) and I just got the answer same as the answer I mentioned at the start since we have to also turn 6+2i into the form re^(i*phi), isn't it????
Write the bold in exact rectangular form, then multiply in rectangular form, as suggested previously.

The question is doing two things:
* Make sure you understand complex rotations.
* Get you to think about rectangular / exponential representation and which is appropriate for the question. The 6+2i is deliberately chosen to be "harder" in exponential form.The

Note you could get there after doing the exponential multiplication, but the trig messing around is just tedious.
Last edited by mqb2766; 2 weeks ago
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#7
(Original post by mqb2766)
Write the bold in exact rectangular form, then multiply in rectangular form, as suggested previously.

The question is doing two things:
* Make sure you understand complex rotations.
* Get you to think about rectangular / exponential representation and which is appropriate for the question. Obviously the answer should be the same, but the 6+2i is deliberately chosen to be "harder" in exponential form.
I will have a try on it. Thank you
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