A regular polygon is formed by a complex equation’s solutions.

[tpupble]

A complex equation has exactly 4 solutions. When joined on the complex plane, these solutions form the vertices of a regular polygon. If the midpoint of one side of the polygon is 1/2 + √3/2 i, determine the complex equation in both polar and Cartesian form.

I have attached my working and diagrams with the full question.

My first step was to plot the midpoint. After that, I plotted the other 3 midpoints, adding or subtracting 1/2 and √3/2 from (0,0) accordingly as shown on my diagram. This is because the 4 solutions would be equidistant from each other, hence the midpoints would also be equidistant from each other.

I then found one solution by using Pythagoras’ theorem to construct a point which would be the same distance from two of the midpoints - (1/2,√3/2) and (-√3/2, 1/2).

Since this point was a solution of the overall complex equation which had 4 roots in total, I took the point to the power of 4 to find the overall equation. Though, I believe part of my Process is faulty as the answer cannot be converted into a polar form with an argument from a special triangle (this question is supposed to be doable without a calculator).

 

[blamocur]

After that, I plotted the other 3 midpoints, adding or subtracting 1/2 and √3/2 from (0,0) accordingly

It looks like you rotated midpoints by 90 degrees, which looks correct.

Since this point was a solution of the overall complex equation which had 4 roots in total, I took the point to the power of 4 to find the overall equation.

I don't get this part. Can you list the 4 vertices of the polygon ?
What is the resulting equation ?

P.S. The problem seems to imply that the center of the polygon is at the origin, but this is not explicitly stated.

 

[blamocur]

Though, I believe part of my Process is faulty as the answer cannot be converted into a polar form with an argument from a special triangle (this question is supposed to be doable without a calculator).

It is doable without a calculator. What is the argument of [imath]\frac{1}{2} + \frac{\sqrt{3}}{2}i[/imath] ?

 

[tpupble]

Sure:

The 4 vertices I got would be
[(1-√3)/2, (1+√3)/2]
[(1+√3)/2, (√3 - 1)/2]
[(1+√3)/2, (-1-√3)/2]
[(-1-√3)/2, (1-√3)/2]

The reason why I took the first solution to the power of 4 is because the goal is to find the 4th roots of the complex number. From what I understand for these complex number root questions, no matter which root I take to its power, I will always yield the same complex equation.

For example: the cube roots of -1 + √3 i are equal to 2^(1/3) cis (2π/9), 2^(1/3) cis (8π/9), 2^(1/3) cis (-4π/9) and no matter which one I take to the power of 3, I will get -1+√3 i.

Therefore, whilst true that I can get the argument of my midpoint without a calculator, I don’t understand why I would need to do so. The midpoint is not one of the solutions to the complex equation I am trying to find - wouldn’t I need my complex equation (which I get from taking one of the roots to the power of 4) to have the ratio which can be obtained from a special triangle?

Also I got 15-8i from taking z = (1-√3)/2 + (1+√3)/2)i to the power of 4.

 

[Dr.Peterson]

A complex equation has exactly 4 solutions. When joined on the complex plane, these solutions form the vertices of a regular polygon. If the midpoint of one side of the polygon is 1/2 + √3/2 i, determine the complex equation in both polar and Cartesian form.

I feel that the problem as stated is incomplete. The solution assumes that the regular polygon (square) is centered at the origin, but the problem doesn't say that. Without that unstated assumption, there are many answers.

One equation that satisfies what is stated is [imath](z - \frac{\sqrt{3}-1}{2}i)^4=1[/imath], whose solutions are

​

The solutions form a square (not centered at the origin), and one midpoint is as given.

I found my equation by starting with [imath]z^4 = k[/imath], taking k=1 to make it easy, and just shifting the graph to put the midpoint where I wanted it.

 

[blamocur]

The 4 vertices I got would be
[(1-√3)/2, (1+√3)/2]
[(1+√3)/2, (√3 - 1)/2]
[(1+√3)/2, (-1-√3)/2]
[(-1-√3)/2, (1-√3)/2]

Looks good to me.

Also I got 15-8i from taking z = (1-√3)/2 + (1+√3)/2)i to the power of 4.

I agree with your approach, but I get a very different number there.

Therefore, whilst true that I can get the argument of my midpoint without a calculator, I don’t understand why I would need to do so.

To get polar representations of the points.