Prove that cis(x) + cis (y) = 2 cos [(x-y)/2]cis[(x+y)/2].

[Masaru]

That is only the first part of the following question:

"Prove that cis(x) + cis (y) = 2 cos [(x-y)/2]cis[(x+y)/2]. Hence show that [(z+1)/(z-1)]⁵ = 1 has solutions of the form z = i cot [n(pi)/5] for n = 1, 2, 3, and 4.

Honestly, this is an extremely hard qustion for me and I cannot even relate the first part of the question to the second one - there seems to be no relation whatsoever....

Anyway, I know that
cis (a + b )
= cos(a+b) +isin(a+b)
= cos(a)cos(b)-sin(a)sin(b)+i[sin(a)cos(b)+sin(b)cos(a)]
= [cos(a) + isin(a)][cos(b) +isin(b)] ......................................edited
= cis(a) * cis(b).

So I did
cis(x) + cis (y)
= cis (x/2 + x/2) + cis (y/2 + y/2)
= cis(x/2)cis(x/2) + cis(y/2)cis(y/2)
= cis²(x/2)+cis²(y/2)
= [cos(x/2)+isin(x/2)]² + [cos(y/2)+isin(y/2)]²
= cos²(x/2) +2isin(x/2)cos(x/2)-sin²(x/2) + cos²(y/2) +2isin(y/2)cos(y/2)-sin²(y/2)

Now I have x/2 and y/2 in every single term, so it is a bit more like the right-hand side of the equation. Having said that, I got stuck and cannot go further than this.

I would much appreciate anyone's help with this qustion.

 

[pka]

That is only the first part of the following question:

"Prove that cis(x) + cis (y) = 2 cos [(x-y)/2]cis[(x+y)/2]. Hence show that [(z+1)/(z-1)]⁵ = 1 has solutions of the form z = i cot [n(pi)/5] for n = 1, 2, 3, and 4.

Is there not typo in the statement? \(\displaystyle c{\Large i}s\left( {\frac{{x - y}}{2}} \right)cis\left( {\frac{{x + y}}{2}} \right)~?\)

 

[Deleted member 4993]

That is only the first part of the following question:

"Prove that cis(x) + cis (y) = 2 cos [(x-y)/2]cis[(x+y)/2]. Hence show that [(z+1)/(z-1)]⁵ = 1 has solutions of the form z = i cot [n(pi)/5] for n = 1, 2, 3, and 4.

Honestly, this is an extremely hard question for me and I cannot even relate the first part of the question to the second one - there seems to be no relation whatsoever....

Anyway, I know that
cis (a + b )
= cos(a+b) +isin(a+b)
= cos(a)cos(b)-sin(a)sin(b)+i[sin(a)cos(b)+sin(b)cos(a)]
= [cos(a) + isin(a)][cos(b) +isin(b)] ......................................edited
= cis(a) * cis(b).

So I did
cis(x) + cis (y)
= cis (x/2 + x/2) + cis (y/2 + y/2)
= cis(x/2)cis(x/2) + cis(y/2)cis(y/2)
= cis²(x/2)+cis²(y/2)
= [cos(x/2)+isin(x/2)]² + [cos(y/2)+isin(y/2)]²
= cos²(x/2) +2isin(x/2)cos(x/2)-sin²(x/2) + cos²(y/2) +2isin(y/2)cos(y/2)-sin²(y/2)

Now I have x/2 and y/2 in every single term, so it is a bit more like the right-hand side of the equation. Having said that, I got stuck and cannot go further than this.

I would much appreciate anyone's help with this question.

Did you try to simplify RHS - by expanding and distributing? If you did what did you get?

 

[Steven G]

I agree with Subhotosh Khan and think that you should expand the rhs

 

[Masaru]

Is there not typo in the statement? \(\displaystyle c{\Large i}s\left( {\frac{{x - y}}{2}} \right)cis\left( {\frac{{x + y}}{2}} \right)~?\)

I have ERRATA for the book with this question in it, which does not say that it is an error. So it is not a typo.

 

[Masaru]

Did you try to simplify RHS - by expanding and distributing? If you did what did you get?

I have expanded the right-hand side of the equation like this, but it seems that I am not getting anywhere...
2cos[(x-y)/2]cis[(x+y)/2]
= 2cos[(x-y)/2]*[cos[(x+y)/2] + isin[(x+y)/2]]
= 2cos[(x+y)/2]*cos[(x-y)/2] + 2isin[(x+y)/2]]*cos[(x-y)/2]
= 2cos(x/2 + y/2)*cos(x/2 - y/2) + 2isin(x/2 + y/2)*cos(x/2 -y/2)
= 2[cos(x/2)cos(y/2) - sin(x/2)sin(y/2)]*[cos(x/2)cos(y/2) +sin(x/2)sin(y/2)] + 2i[(sin(x/2)cos(y/2) + cos(x/2)sin(y/2))*(cos(x/2)cos(y/2) + sin(x/2)sin(y/2))]
= 2[(cos(x/2)cos(y/2))²-(sin(x/2)sin(y/2))²] + 2i[sin(x/2)cos(x/2)cos²(y/2) + sin²(x/2)sin(y/2)cos(y/2) + sin(y/2)cos(y/2)cos²(x/2) + sin(x/2)sin²(y/2)cos(x/2)]

And I have no idea how I can simplify this lengthy expansion to make it equal to cis x + cis y.

Is anything wrong with this approach and/or my working above?

 

[Dr.Peterson]

I would consider using the product-to-sum identities:

sin(x)cos(y) = [sin(x+y) + sin(x-y)]/2​

cos(x)sin(y) = [sin(x+y) - sin(x-y)]/2​

cos(x)cos(y) = [cos(x-y) + cos(x+y)]/2​

sin(x)sin(y) = [cos(x-y) - cos(x+y)]/2​

Applying a couple of those after your third line should do the trick.

 

[Masaru]

I have expanded the right-hand side of the equation like this, but it seems that I am not getting anywhere...
2cos[(x-y)/2]cis[(x+y)/2]
= 2cos[(x-y)/2]*[cos[(x+y)/2] + isin[(x+y)/2]]
= 2cos[(x+y)/2]*cos[(x-y)/2] + 2isin[(x+y)/2]]*cos[(x-y)/2]
= 2cos(x/2 + y/2)*cos(x/2 - y/2) + 2isin(x/2 + y/2)*cos(x/2 -y/2)
= 2[cos(x/2)cos(y/2) - sin(x/2)sin(y/2)]*[cos(x/2)cos(y/2) +sin(x/2)sin(y/2)] + 2i[(sin(x/2)cos(y/2) + cos(x/2)sin(y/2))*(cos(x/2)cos(y/2) + sin(x/2)sin(y/2))]
= 2[(cos(x/2)cos(y/2))²-(sin(x/2)sin(y/2))²] + 2i[sin(x/2)cos(x/2)cos²(y/2) + sin²(x/2)sin(y/2)cos(y/2) + sin(y/2)cos(y/2)cos²(x/2) + sin(x/2)sin²(y/2)cos(x/2)]

And I have no idea how I can simplify this lengthy expansion to make it equal to cis x + cis y.

Is anything wrong with this approach and/or my working above?

Thank you very much for your help, Dr. Peterson.

With your help, I was able to finally work it out as follows:

2cos[(x-y)/2]cis[(x+y)/2]
= 2cos[(x-y)/2]*[cos[(x+y)/2] + isin[(x+y)/2]]
= 2cos[(x+y)/2]*cos[(x-y)/2] + i2sin[(x+y)/2]]*cos[(x-y)/2]
= cos[(x+y)/2 + (x-y)/2] + cos[(x+y)/2 - (x-y)/2] + i [sin[(x+y)/2 + (x-y)/2] + sin [(x+y)/2 - (x-y)/2]]
= cos(x) + cos(y) + i[sin(x) + sin(y)]
= [cos(x) + isin(x)] + [cos(y) + isin(y)]
= cis(x) + cis(y)
= LHS

But I still cannot figure out the second part of the question that says:
Hence show that [(z+1)/(z-1)]⁵ = 1 has solutions of the form z = i cot [n(pi)/5] for n = 1, 2, 3, and 4.

The qustion has the word "Hence", which indicates that we need to use the first part of the question, cis(x) + cis (y) = 2 cos [(x-y)/2]cis[(x+y)/2], for the second part.

However, I cannot relate the first part to the second one in any way whatsoever.

I would much appreciate it if you can give me a clue on how we can possibly utilise the first part for the second one.

 

[Masaru]

With regard to the second part of the question, the following is all that I can do:

[(z +1)/(z -1)]⁵ = 1 = cis [0 + n2π] where n = 0, 1, 2, 3....
So, (z + 1)/(z -1) = cis [(1/5)*(0 + n2π)] = cis (n2π/5)
So, z + 1 = (z - 1)*cis (n2π/5)
So, z - z*cis (n2π/5) = - 1 - cis (n2π/5)
So, z [1 - cis(n2π/5)] = - 1 - cis (n2π/5)
Thus, z = [cis (n2π/5) +1]/[cis (n2π/5) - 1]

What I have found so difficult with this question is that the equation in the first part of the question has two unknowns, namely x and y, which are the angle sizes, even though the second part has only one unknown "z", which is a complex number.

So I do not have any idea how I can use the first part for the second one.

I would much appreciate it if there is someone out there who can help me to see a connection between the first part and the second part of the question.

 

[Dr.Peterson]

There are actually 5 5th roots, so your work is incomplete.

But I haven't been able to figure out what they want you to do. I agree that it seems entirely unrelated to the first part.

 

[Masaru]

There are actually 5 5th roots, so your work is incomplete.

But I haven't been able to figure out what they want you to do. I agree that it seems entirely unrelated to the first part.

Do you mean that "n" can take 0?
Is that the reason why you think that there are 5 roots, not 4?
But the qustion that I have here clearly says that "n" takes only 1, 2, 3, 4. (and not 0)
I do not think that "n" can take 0 because, as you can see from my incomplete working above, z= [cis (n2π/5) +1] / [cis (n2π/5) - 1], if we replace "n" with 0, then the denominator becomes cis(0) - 1 = 1 - 1 = 0, which is unacceptable.

From z = [cis (n2π/5) +1] / [cis (n2π/5) - 1], I have further changed it to:
z = [cis (n2π/5) +cis (0)] / [cis (n2π/5) + cis (π)], which can be also expressed as [cis (n2π/5) - i^2}] / [cis (n2π/5) + i^2]

If we can somehow remove the denominator or make it free from "i", then it will be the form of cis (x) + cis (y) to get to the solution.

But I have not been able to figure out how to do it because we have cis (π) in the denominator.

If we multiply both top and bottom of the fraction by cis (n3π/5), for example, we get:
z = [cis (n2π/5) +cis (0)]*cis (n3π/5) / [cis (n2π/5) + cis (π)]*cis (n3π/5)
= [cis (n2π/5 + n3π/5) +cis (0 + n3π/5)] / [cis (n2π/5 +n3π/5 ) + cis (π + n3π/5)]
= [ cis (nπ) + cis (n3π/5) ] / [cis (nπ) + cis (π + n3π/5)]
= [ cis (nπ) + cis (n3π/5) ] / [{cis (π)}ⁿ + cis (π + n3π/5)]
= [ cis (nπ) + cis (n3π/5) ] / [( - 1 )ⁿ + cis (π + n3π/5)]

So it does not work.

 

[Dr.Peterson]

Actually, I misread what you wrote. I didn't mean to say n could be 0 in the problem.

 

[Masaru]

Actually, I misread what you wrote. I didn't mean to say n could be 0 in the problem.

I have made some progress as follows, however, what I have ended up with is slightly different to icot(nπ/5) - the only difference is that I've got " - " sign in front of icot(nπ/5) - even though I have checked my working below many times and there seems to be no error :

[(z +1)/(z -1)]⁵ = 1 = cis [0 + n2π] where n = 0, 1, 2, 3....
So, (z + 1)/(z -1) = cis [(1/5)*(0 + n2π)] = cis (n2π/5)
So, z + 1 = (z - 1)*cis (n2π/5)
So, z - z*cis (n2π/5) = - 1 - cis (n2π/5)
So, z [1 - cis(n2π/5)] = - 1 - cis (n2π/5)
So, z = [cis (n2π/5) +1]/[cis (n2π/5) - 1]
= [cis (n2π/5) + cis (0)] / [cis (n2π/5) + cis (π)]

Then using the first part of the question,

= [2cos{(2nπ/5 - 0)/2}*cis{(2nπ/5 + 0)/2}]/[2cos{(2nπ/5 - π)/2}*cis{(2nπ/5 + π)/2}]
= [cos(nπ/5)cis(nπ/5)] / [cos(nπ/5 - π/2 )cis(nπ/5 + π/2)]
= [cos(nπ/5)cis(nπ/5)] / [cos(π/2 - nπ/5)cis(nπ/5 + π/2)]
= [cos(nπ/5)/cos(π/2 - nπ/5)][cis(nπ/5)/cis(nπ/5 + π/2)]
= [cos(nπ/5)/sin(nπ/5)][cis(nπ/5 - (nπ/5 + π/2)]
= cot(nπ/5)cis( - π/2)
= cot(nπ/5)[cos( - π/2)+ isin( - π/2)]
= cot(nπ/5)[cos(π/2) - isin(π/2)]
= cot(nπ/5)[0 - i]
= - icot(nπ/5)

Would you be able to tell me something wrong with my working above or the reason why I've got this negative sign, which we need to somehow get rid of in order to make it consistent with what the second part of the question says?

 

[Masaru]

Actually, I misread what you wrote. I didn't mean to say n could be 0 in the problem.

I have finally worked out the second part of the qustion as follows:

[(z +1)/(z -1)]⁵ = 1 = cis [0 + n2π] where n = 0, 1, 2, 3....
So, (z + 1)/(z -1) = cis [(1/5)*(0 + n2π)] = cis (n2π/5)
So, z + 1 = (z - 1)*cis (n2π/5)
So, z - z*cis (n2π/5) = - 1 - cis (n2π/5)
So, z [1 - cis(n2π/5)] = - 1 - cis (n2π/5)
So, z = [cis (n2π/5) +1]/[cis (n2π/5) - 1]
= [cis (n2π/5) + cis (0)] / [cis (n2π/5) + cis (π)]

Then using the first part of the question,

= [2cos{(2nπ/5 - 0)/2}*cis{(2nπ/5 + 0)/2}]/[2cos{(2nπ/5 - π)/2}*cis{(2nπ/5 + π)/2}]
= [cos(nπ/5)cis(nπ/5)] / [cos(nπ/5 - π/2 )cis(nπ/5 + π/2)]
= [cos(nπ/5)cis(nπ/5)] / [cos(π/2 - nπ/5)cis(nπ/5 + π/2)]
= [cos(nπ/5)/cos(π/2 - nπ/5)][cis(nπ/5)/cis(nπ/5 + π/2)]
= [cos(nπ/5)/sin(nπ/5)][cis(nπ/5 - (nπ/5 + π/2)]
= cot(nπ/5)cis( - π/2)
= cot(nπ/5)[cos( - π/2)+ isin( - π/2)]
= cot(nπ/5)[cos(π/2) - isin(π/2)]
= cot(nπ/5)[0 - i]
= - icot(nπ/5)

Then:

= i[- cot(nπ/5)]
= icot(π - nπ/5)
= icot[(5 - n)π/5]
= icot(mπ/5) where m = 5 - n and n = 1, 2, 3, 4
= icot(nπ/5) where n = 5 - m and m = 1, 2, 3, 4 (i.e. n = 1, 2, 3, 4)

Note that:
1) n cannot take 0 because cot(0) = cos(0)/sin(0) and sin(0) = 0. The denominator cannot be 0.
2) n cannot take a multiple of 5, either, because again sin[(5 - n)π/5] = 0 if n is a multiple of 5.
3) The period of cot is π, so for n = 6, 7, 8, 9 (11, 12, 13, 14...) the value of cot(nπ/5) would be exactly the same as the one for 1, 2, 3, 4.

Therefore, [(z+1)/(z-1)]⁵ = 1 has solutions of the form z = i cot [nπ/5] for n = 1, 2, 3, and 4.

Please let me know if there is anything wrong with my understanding or working above.

Thank you.

 

[Dr.Peterson]

Beautiful! I didn't have time yesterday to examine your work closely; what you have now found is that {- icot(nπ/5) , n = 1, 2, 3, 4} is the same set as {icot(nπ/5) , n = 1, 2, 3, 4}, because the elements are just in a different order; so your initial answer was not really wrong.