Help Evaluating Complex Numbers

In summary, the conversation is about evaluating the product of sines, sin(pi/7) * sin(2pi/7) * sin(3pi/7), and using the cosine product to find the solution. The conversation discusses finding the solution using complex numbers and the formula (z^7-1)/(z-1) = prod(1-zeta^k). It is suggested to try using complex exponentials and substituting z=1 to simplify the problem.
  • #1
VertexOperator
79
0

Homework Statement



Evaluate:

Homework Equations



[tex]sin\frac{\pi }{7}.sin\frac{2\pi }{7}.sin\frac{3\pi }{7}[/tex]

The Attempt at a Solution



[tex]Using[/tex] [tex]z^{7}-1[/tex] [tex]got:[/tex]
[tex]cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex]
 
Last edited:
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  • #3
micromass said:
Seems correct!

But my solution is for the cosine product. How can I find the solution to the question from what I got?
 
  • #4
Can't you repeat much of the same proof? Where does it fail?
 
  • #5
The way I got: [tex]cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex] is by solving the equation [tex]z^{7}-1=0[/tex]
[tex]z=1,cis\pm\frac{2\pi }{7},cis\pm\frac{4\pi }{7},cis\pm\frac{6\pi }{7}[/tex]
[tex]w+w^{-1}=cis\frac{2\pi }{7}+cis\frac{-2\pi }{7}=2cos\frac{2\pi }{7}[/tex]
Similarly;
[tex]w^{2}+w^{-2}=-2cos\frac{3\pi }{7}[/tex]
[tex]w^{3}+w^{-3}=-2cos\frac{\pi }{7}[/tex]
[tex](w+w^{-1})(w^{2}+w^{-2})(w^{3}+w^{-3})=[/tex]
[tex]w^{6}+1+w^{2}+w^{3}+w^{4}+w^{5}+1+w=1[/tex]
, since [tex]w^{6}+w^{5}+w^{4}+w^{3}+w^{2}+w+1=0 \therefore cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex]
So I don't think I can do the same to get the produce for sine.
 
  • #6
It might be easier to determine

[tex]\prod_{k=0}^6 \sin(k\pi/7)[/tex]

Do this by expressing the sine as complex exponentials. Also try to make use of the following formula

[tex]\frac{z^7-1}{z-1}=\prod_{k=1}^6 (1- \zeta^k)[/tex]

with [itex]\zeta = \cos(\pi/7) + i\sin(\pi/7)[/itex]. Let z go to 1.
 

1. What are complex numbers?

Complex numbers are numbers that consist of a real part and an imaginary part. They are typically written in the form a + bi, where a is the real part and bi is the imaginary part, with i representing the imaginary unit √-1.

2. How do you add complex numbers?

To add complex numbers, you simply add the real parts together and the imaginary parts together. For example, (3 + 2i) + (5 + 4i) = (3+5) + (2i+4i) = 8 + 6i.

3. What is the conjugate of a complex number?

The conjugate of a complex number a + bi is the number a - bi, where the sign of the imaginary part is changed. For example, the conjugate of 3 + 2i is 3 - 2i.

4. How do you multiply complex numbers?

To multiply complex numbers, you use the FOIL method, just like when multiplying binomials. For example, (3 + 2i)(5 + 4i) = 15 + 12i + 10i + 8i^2 = 15 + 22i - 8 = 7 + 22i.

5. What is the modulus of a complex number?

The modulus (or absolute value) of a complex number a + bi is the distance from the origin to the point on the complex plane represented by that number. It is calculated using the Pythagorean theorem: |a + bi| = √(a^2 + b^2).

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