| Veronica
Fontama |
Happy New Year to Everyone, I will be grateful if someone can help provide a solution and/or explain the solution provided by for Question 10 in Advanced Problems in Mathematics by Dr. Siklos. Here is the question: Explain the geometrical relationship between the points in the Argand diagram represented by the complex numbers z and (z-a)eiq. (i) Write down the necessary and sufficient conditions that thedistinct complex numbers a, b and g represent the vertices of an equilateral triangle taken in clockwise order. Show that a, b and g represent the vertices of an equilateral triangle if and only if a2+b2+g2-bg-g a-ab = 0 (ii) Find the necessaryand sufficient conditions on the complex numbersa,b and c for the roots of the equation z3+a z2+b z+c=0 to lie at the vertices of an equilateral triangle in the Argand diagram. I have looked at the solution provided by Dr. Siklos but it is not quite clear to me. If someone can break each step and explain further it will probably help. Thanks so much for the help. Veronica |
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| Colin
Prue |
Does this sudden interest in STEP mean that you have received an offer then veronica? for the first part, draw yourself vectors a and z on the complex plane. turn these into a parallelogram with side z and diameter a, in order to convince yourself that the remaining side can be represented by a-z. In other words, z is |a-z| away from a. Therefore we can now describe z as a-(a-z), or a+(z-a) with an understanding of the geometry. Are you familiar with the description of a complex number in the following way: r(cos x+i sin x)=r ei x then by the rule, multiply the moduli, add the arguments, (z-a)ei x is an anti-clockwise rotation of (z-a) by x rad. hence 'a+(z-a)ei x is a rotation of z by x rad about a.' as for part i), you could require that a, b, g can be described by; a z a+(z-a)ei x i.e. (a-z) is a p/3 rotation of a-[(z-a)ei x]. i will leave it to you to express this in terms of a, b, g, except for mentioning that multiplying by a complex number modulus 1, rotates but does not lengthen for the second part of i), it is necessary to make sure that generality is not lost - that is, it must not be important which vertices are a, b, g. therefore you need to consider both the clockwise and anticlockwise triangles. for this you need to have found the correct expression for the previous paragraph, so i will wait until you reply to show that you understand, or do not, what i have said up until now, hinting only that if you know: x=0 or y=0 Þ x y=0 multiply out and you will attain the desired expressions |
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| Anuj
Shah |
This diagram might help:
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| Anuj
Shah |
There is a slight mistake, the angle labelled q at the bottom is meant to be between (z-a)eiq and z-a |
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| Colin
Prue |
anuj, i do not understand why the angle you have labelled is q, rather i understood it to be only the angle at a that was q on a point of clarification: in my above post x was substituted for every q because of ease |
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| Anuj
Shah |
Colin, have you read the part above where i said i made a mistake. if no, read it, everything should make sense. if yes, the two marked angles should be the same and equal to q, (bearing in mind there is a mistake in the diagram) because (z-a)eiq is parallel with the verticalish dotted line and z-a is parallel with horizontalish dotted line. |
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| Veronica
Fontama |
Dear Colin and Anuj, Thank you for responding to my question. Sorry I wasn't able to get back to you early enough. For Colin, the answer is no, I have not been offered a place. I wish I was young again to go back and study for another degree. My interest in step is because I am helping a student who might be offered a place. For your responses, I revised my vector algebra and I am happy to tell you now that I do understand the geometric relationship between those complex numbers. The Figure was also very helpful. My first question however, is the reasoning behind the assumptions in part (i). That is: Letting a, b and g to be described by: a z a+(z-a)eip/e I presume this assumption if from the first part of he question. Why was it obvious that the required conditions for (i) can be derived from the expression involving a, z and (z-a)eiq Did this come from observing the Figure and understanding that z, a, and z-a will represent the sides of an equilateral triangle. If this is the case, I can see that: (g-a)=(b-a)eip/3 Is this the expression you expect? Veronica |
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| Colin
Prue |
That was exactly the intention, as we know that the triangle will be isoceles (as seen in Anuj's figure) - we merely have to be cunning in our choice of angle to make it the special isoceles that is also equilateral. And yes indeed, that was the result i wanted. And so for the second part:...perhaps i'll wait until tomorrow when my brain's online again |