Modulus of the difference of two complex numbers

In summary, the conversation is about using geometry to show that |z3-z-3| = 2sin3θ for z=cisθ, 0<θ<∏/6. The conversation includes a discussion of using the definition of "cis" and manipulating it to show that z^3 - z^-3 = 2sin3θ. There is also a mention of using geometry to solve the problem, specifically by drawing a rhombus and determining the length of its diagonal, which is equivalent to 2sin3θ. The conversation ends with a clarification that the factor of i does not change the answer as the question is asking for the modulus.
  • #1
Fourthkind
5
0
Hi guys,
I've been trying to help a friend with something that I learned in class but I'm now finding it hard to solve myself. The problem goes as follows:

Use geometry to show that |z3-z-3| = 2sin3θ

For z=cisθ, 0<θ<∏/6

Now, I chose ∏/12 as my angle and plotted all this on an Argand diagram, but I don't know how to prove that |z3-z-3| = 2sin3θ in such general form. I know that it is right, but I don't know how to get there. Right now I have a isosceles triangle with two lengths known and two angles known (both 3θ). I know the triangle is right-angled but I don't know if I can use this information since that is specific to my chosen angle, right?

I've spent quite some time on it and I'd really like to understand it. So far I've got xsin3θ=1 where x = |z3-z-3|.

Please help!
 
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  • #2
Fourthkind said:
Hi guys,
I've been trying to help a friend with something that I learned in class but I'm now finding it hard to solve myself. The problem goes as follows:

Use geometry to show that |z3-z-3| = 2sin3θ

For z=cisθ, 0<θ<∏/6 if [itex]z= cis(\theta)[/itex] then [itex]z^3= cis(3\theta)[/itex] and [itex]z^{-3}= cis(-3\theta)[/tex]
Do you understand what "cis" means? This takes about one line from that. You don't need to do anything at all with "Argand diagrams". If z= cis(θ) then [itex]z^3= cis(3\theta)[/itex] and [itex]z^{-3}= cis(-3\theta)[/itex] so that [itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)[/itex]. Write that out in terms of the definition of "ciz".

Now, I chose ∏/12 as my angle and plotted all this on an Argand diagram, but I don't know how to prove that |z3-z-3| = 2sin3θ in such general form. I know that it is right, but I don't know how to get there. Right now I have a isosceles triangle with two lengths known and two angles known (both 3θ). I know the triangle is right-angled but I don't know if I can use this information since that is specific to my chosen angle, right?

I've spent quite some time on it and I'd really like to understand it. So far I've got xsin3θ=1 where x = |z3-z-3|.

Please help!
 
  • #3
HallsofIvy said:
Do you understand what "cis" means? This takes about one line from that. You don't need to do anything at all with "Argand diagrams". If z= cis(θ) then [itex]z^3= cis(3\theta)[/itex] and [itex]z^{-3}= cis(-3\theta)[/itex] so that [itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)[/itex]. Write that out in terms of the definition of "ciz".

Could you explain this a little more?

I managed to solve it by drawing a rhombus with |[itex]z^3- z^{-3}[/itex]|being the vertical length along the imaginary axis from the origin to the point [itex]z^3- z^{-3}[/itex] (by definition) then drawing a perpendicular bisector (that is, parallel to the real axis) which created four right-angled triangles with their 90° angles joined to the midpoint between the origin and point [itex]z^3- z^{-3}[/itex]. Using one of these triangles, I determined the length of OM was [itex]sin3θ[/itex] and therefore O to [itex]z^3- z^{-3}[/itex] was twice that since M is the midpoint. I solved it like this since the problem stated it needed to be solved using geometry specifically.

I'll attach a picture to show you what I mean. It doesn't show in it but, line OA is equal in length to [itex]z^{3}[/itex]
 

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  • #4
You didn't say "the problem stated it needed to be solved using geometry specifically"!

My idea was [itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)= cos(3\theta)+ sin(3\theta)- (cos(-3\theta)+ sin(-3\theta)[/itex][itex]= cos(3\theta)+ sin(3\theta)- cos(3\theta)+ sin(3\theta)= 2sin(3\theta)[/itex].
 
  • #5
Fourthkind said:
Use geometry to show that |z3-z-3| = 2sin3θ

So I assumed I had to do it in a way similar to the way I did.

HallsofIvy said:
My idea was [itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)= cos(3\theta)+ sin(3\theta)- (cos(-3\theta)+ sin(-3\theta)[/itex][itex]= cos(3\theta)+ sin(3\theta)- cos(3\theta)+ sin(3\theta)= 2sin(3\theta)[/itex].

This is probably stupid but when I do it that way, I end up with [itex]2isin3θ[/itex] since:
[itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)= cos(3\theta)+ isin(3\theta)- (cos(-3\theta)+ isin(-3\theta)[/itex][itex]= cos(3\theta)+ isin(3\theta)- cos(3\theta)+ isin(3\theta)= 2isin(3\theta)[/itex]

My class just recently started our complex number topic and so my knowledge of [itex]cis[/itex] is extremely basic. Should I not be writing [itex]z^3[/itex] as [itex]cos3θ+isin3θ[/itex].

Thanks for your input.
 
  • #6
Fourthkind said:
This is probably stupid but when I do it that way, I end up with [itex]2isin3θ[/itex] since:
[itex]z^3- z^{-3}= cis(3\theta)- cis(-3\theta)= cos(3\theta)+ isin(3\theta)- (cos(-3\theta)+ isin(-3\theta)[/itex][itex]= cos(3\theta)+ isin(3\theta)- cos(3\theta)+ isin(3\theta)= 2isin(3\theta)[/itex]

The question is asking for the modulus so a factor of i does not change the answer.

Edit: You asked for a geometric reason, so plot any z with the given constraint. z^3 will have the same magnitude (1) and will make an angle 3θ with the real axis if z makes and angle θ. z^-3 will just be the reflection of z^3 about the real axis. The length |z^3-z^-3| is just the length of the line connecting these two points so sin(3θ) will be half of this.
 
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What is the modulus of the difference of two complex numbers?

The modulus of the difference of two complex numbers is a measure of the distance between the two numbers on the complex plane. It is calculated by taking the absolute value of the difference between the two complex numbers.

How is the modulus of the difference of two complex numbers related to the Pythagorean theorem?

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. In the complex plane, the modulus of the difference of two complex numbers can be thought of as the hypotenuse of a right triangle, with the real and imaginary parts of the numbers as the other two sides.

Can the modulus of the difference of two complex numbers be negative?

No, the modulus of the difference of two complex numbers is always a positive value. This is because it represents a distance, which cannot be negative.

How is the modulus of the difference of two complex numbers used in engineering and physics?

In engineering and physics, the modulus of the difference of two complex numbers is often used to calculate the magnitude or amplitude of a wave. It is also used in vector calculations, such as finding the resultant force of two vectors.

What is the significance of the modulus of the difference of two complex numbers in the study of complex analysis?

In complex analysis, the modulus of the difference of two complex numbers is used to define the concept of continuity and differentiability. It is also used to study the behavior of complex functions and their graphs on the complex plane.

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