Complex number questions (1 Viewer)

[poptarts12345]

1. Prove that if z1+z2+z3= 0 and|z1|=|z2|=|z3|= 1, then the pointsz1, z2andz3are the vertices of an equilateral triangle inscribed in the unit circle.

2. Let a,b,c be complex numbers representing vertices of triangle ABC, let w = cos(2 pi/3) + isin (2pi/3) show that triangle ABC is equilateral if and only if a+bw+cw^2=0

3. Prove that if the origin, z1, z2 and z3 are concyclic, then 1/(z1), 1/(z2), 1/(z3) are collinear.

4. Sketch the locus of z if 1≤|z| ≤ 2 and pi/3 ≤ arg(z) ≤ (2pi)/3

 

[fan96]

Here's a solution for Q1.

If and , then for any we also have
and .
Therefore, without loss of generality, let .

Write and . Then





The RHS has magnitude 1, so



Thus

Since the LHS is real, the RHS must be real too. Hence it is either 1 or -1, so


Thus has principal value 0, so and are conjugates.

This is enough information to conclude that and are the points that form an equilaterial triangle with the point .

 

[tickboom]

For Q1, I think an alternative approach could be to use roots of unity. For example, consider the solutions to ... Plotting these in the unit circle would give rise to an equilateral triangle, each solution would have modulus of 1, and the sum of the roots would add up to 0 ... For some reason this approach is what jump out to be as being most intuitive.

 

[Trebla]

For Q1, I think an alternative approach could be to use roots of unity. For example, consider the solutions to ... Plotting these in the unit circle would give rise to an equilateral triangle, each solution would have modulus of 1, and the sum of the roots would add up to 0 ... For some reason this approach is what jump out to be as being most intuitive.

View attachment 29712

Isn’t this a proof of the converse statement? You are given the sum is zero and modulus is 1. You will then need to effectively show that they satisfy the cubic roots of unity, rather than the other way around.

 

[tickboom]

Isn’t this a proof of the converse statement? You are given the sum is zero and modulus is 1. You will then need to effectively show that they satisfy the cubic roots of unity, rather than the other way around.

Yes good point. Please disregard my comments

 

[Drdusk]

1. Prove that if z1+z2+z3= 0 and|z1|=|z2|=|z3|= 1, then the pointsz1, z2andz3are the vertices of an equilateral triangle inscribed in the unit circle.

2. Let a,b,c be complex numbers representing vertices of triangle ABC, let w = cos(2 pi/3) + isin (2pi/3) show that triangle ABC is equilateral if and only if a+bw+cw^2=0

3. Prove that if the origin, z1, z2 and z3 are concyclic, then 1/(z1), 1/(z2), 1/(z3) are collinear.

4. Sketch the locus of z if 1≤|z| ≤ 2 and pi/3 ≤ arg(z) ≤ (2pi)/3

For part 2:

 

[fan96]

For Q3:

Let describe a circle on the complex plane with centre .
Substitute , and expand to get









If the circle intersects the origin then , and so



which represents a line in when is fixed.

(If you're having trouble seeing this, take the line and multiply it by . Complex number multiplication is a composition of scaling and rotation, which preserves lines. The resulting line satisfies .)

This proves that the map sends any origin-intersecting circle to a line, which is actually a special case of a Mobius transformation.

Your Q4 is simply the intersection of a circle and a sector.

 

[Nav123]

For part 3: Unfortunately circle geometry has been removed from the syllabus but here is a circle geometry proof anyways:
We essentially have to prove: As the two vectors must have the same direction.

Plotting the information in the question:



Using vector addition it immediately follows that and . Furthermore: and

Since The exterior angle of a cylic quadilateral equals the interior opposite angle . In other words:
.

Now it is simply a matter of working backwords to get to what is required.

.

.

.

 

[CM_Tutor]

For part 3: Unfortunately circle geometry has been removed from the syllabus but here is a circle geometry proof anyways:
We essentially have to prove: As the two vectors must have the same direction.

Plotting the information in the question:

View attachment 29713

Using vector addition it immediately follows that and . Furthermore: and

Since The exterior angle of a cylic quadilateral equals the interior opposite angle . In other words:
.

Now it is simply a matter of working backwords to get to what is required.

.

.

.

First, the lack of circle geometry in the MX2 syllabus is a design flaw given the number of complex number problems that are most naturally viewed from that perspective. I don't see how an MX2 student can be well-prepared for MX2 MCQ and complex number sketching without learning some basic circle geometry theorems.

Secondly, I like the approach but I think it is incomplete. Nav123 writes that we "essentially have to prove



as the two vectors must have the same direction. However, unless I am missing something, this does not necessarily have to be true. We need the points , , and to be collinear. If lies between and then Nav123's goal is sufficient to prove that the points are collinear as the two vectors must point in the same direction and therefore must lie on a single line as they share a common point... but, do the reciprocals have to lie along the common line ordered so that is between the others? I don't see why that is necessarily the case, and if does not lie between and then the arguments of the stated vectors must differ by an odd multiple of and the principal arguments must be opposite. In that case, we would need to prove



or an equivalent statement, such as that



as we know that the direction to must be the same from and .

A complete proof would either need to cover all possibilities (by taking cases), or it must be set up in such a way that the placement of the three points on the circle (relative to the origin) allows it to be know which reciprocal will be in the centre of the line, but doing so without any loss of generality. Without any preamble being included, this seems to me to be a problem where drawing the diagram can lead to an unnoticed assumption being made and thus restricting the proof to dealing only with a single case. Or am I missing something?

Thirdly, what do others think about how such answers would be taken in school assessments and the HSC. They certainly are valid, but I can see an argument that credit should be restricted when invoking theory not covered in the course, unless a proof of the result being invoked is included. Put another way, I can't see any grounds for less than full marks for a valid proof that invokes a theorem from outside the syllabus and proves that theorem is true as part of the answer. But, what if the result is simply asserted? Assuming there is nothing in the question to prevent its use (like "by using method X", etc). How would HSC markers see that? Schools are free to take their own perspectives, but what do we think those might be? Thoughts? Thanks.

 

[Nav123]

First, the lack of circle geometry in the MX2 syllabus is a design flaw given the number of complex number problems that are most naturally viewed from that perspective. I don't see how an MX2 student can be well-prepared for MX2 MCQ and complex number sketching without learning some basic circle geometry theorems.

Secondly, I like the approach but I think it is incomplete. Nav123 writes that we "essentially have to prove



as the two vectors must have the same direction. However, unless I am missing something, this does not necessarily have to be true. We need the points , , and to be collinear. If lies between and then Nav123's goal is sufficient to prove that the points are collinear as the two vectors must point in the same direction and therefore must lie on a single line as they share a common point... but, do the reciprocals have to lie along the common line ordered so that is between the others? I don't see why that is necessarily the case, and if does not lie between and then the arguments of the stated vectors must differ by an odd multiple of and the principal arguments must be opposite. In that case, we would need to prove



or an equivalent statement, such as that



as we know that the direction to must be the same from and .

A complete proof would either need to cover all possibilities (by taking cases), or it must be set up in such a way that the placement of the three points on the circle (relative to the origin) allows it to be know which reciprocal will be in the centre of the line, but doing so without any loss of generality. Without any preamble being included, this seems to me to be a problem where drawing the diagram can lead to an unnoticed assumption being made and thus restricting the proof to dealing only with a single case. Or am I missing something?

Thirdly, what do others think about how such answers would be taken in school assessments and the HSC. They certainly are valid, but I can see an argument that credit should be restricted when invoking theory not covered in the course, unless a proof of the result being invoked is included. Put another way, I can't see any grounds for less than full marks for a valid proof that invokes a theorem from outside the syllabus and proves that theorem is true as part of the answer. But, what if the result is simply asserted? Assuming there is nothing in the question to prevent its use (like "by using method X", etc). How would HSC markers see that? Schools are free to take their own perspectives, but what do we think those might be? Thoughts? Thanks.

Yes I guess I should have clarified that this same argument can be repeated for different orders of on the circle, but I assumed by the symmetry it is quite easy to see how.