Surjectivity of Sin/Cos over the Complex Plane

In summary, in order to verify the identities Cos(z) = (exp(iz)+exp(-iz))/2 and sin(z) = (exp(iz)-exp(-iz))/2i, it is necessary to use Euler's formula and the even/odd nature of cos/sin. To show that sin(z) and cos(z) are surjective from C to C, one can use these identities to find an explicit inverse for sin and cos. By solving for z, it is possible to determine all values of z in C such that sin(z) = 12i/5. However, it is important to note that there are an infinite number of solutions, as the exponential and logarithmic functions are periodic over the complex numbers
  • #1
moo5003
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Verify Identities Cos(z) = (exp(iz)+exp(-iz))/2 sin(z)= (exp(iz)-exp(-iz))/2i

Using identities show that sin(z) and cos(z) are surjective from C to C.

Determine all z in C such that sin(z) = 12i/5

Solutions: The verification seems simple using eulers formula and the even/odd nature of cos/sin. My problem is using the identities to show sin/cos are surjective.

Attempt:
Given m in C, m = cos(z) = (exp(iz)+exp(-iz))/2 implying 2m = exp(iz)+exp(-iz) implying Log(2m) = Log(exp(iz)(1+exp(-2iz)) = izLog(1+exp(-2iz). From here I was going to expand the log into a taylor series.. though I'm not sure how that will help. Any suggestions would be appreciated. Obviously, once I can find some equation to determine z, I can just plug it into find z for 12i/5 and then add steps of 2ipi since Log is periodic over the complex numbers.

Just spent around 2 hours trying to tackle the problem from a different perspective... not sure how to proceed. Any help, if just a possible hint from someone who is unsure themselves would be appreciated.
 
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  • #2
Alright, I've made some progress. Given m in C, m = cos(z) =
1/2[e^iz + e^-iz] implying 2m = e^-iz[1+e^2iz] implying
2me^iz = 1 + e^2iz implying e^2iz -2me^iz +1 = 0 implying
e^iz = 2m +or- Sqrt(4m^2 - 4) / 2 = m+or- Sqrt(m^2 - 1)
Thus: z = -iLog[m +or- (m^2-1)^(1/2)]mod2pi

Problem: When I did the same thing for sin(z) I got the same answer... Is that correct? Furthermore, the last part of the problem is to find all z such that m = 12i/5.

sin(z) = 12i/5 what are possible values of z? I'm a little unsure how to proceed... or at least give my answer.

I could just leave it as z = -iLog(12i/5 +or- iSqrt(119/25)) Though I assume there is a way to simplify it more.
 
  • #3
moo5003 said:
Alright, I've made some progress. Given m in C, m = cos(z) =
1/2[e^iz + e^-iz] implying 2m = e^-iz[1+e^2iz] implying
2me^iz = 1 + e^2iz implying e^2iz -2me^iz +1 = 0 implying
e^iz = 2m +or- Sqrt(4m^2 - 4) / 2 = m+or- Sqrt(m^2 - 1)
Thus: z = -iLog[m +or- (m^2-1)^(1/2)]mod2pi

Problem: When I did the same thing for sin(z) I got the same answer... Is that correct? Furthermore, the last part of the problem is to find all z such that m = 12i/5.

sin(z) = 12i/5 what are possible values of z? I'm a little unsure how to proceed... or at least give my answer.

I could just leave it as z = -iLog(12i/5 +or- iSqrt(119/25)) Though I assume there is a way to simplify it more.
What you are saying is that you have absolutely no idea what the problem is asking!

Apparently you are trying to solve for values of v that make cos(z)= (1/2)(eiz- eiz) true. The problem asked you to verify that it is an identity- that is, that it is true for all z.

Start by writing out Taylor series for cos(z), eiz, and e-iz.
 
  • #4
No... He says he understands the identities. What he is trying to do is construct an explicit inverse for sin and cos (ie given m=sin(z) find a value for z). He wants to show they are surjective.
 
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  • #5
I think you are essentially getting it. You wind up with quadratic equations for e^(iz) and then you finally get z with a log. Be careful though, you aren't going to get only two solutions for z. exp(i*z*(1+2*pi*n))=exp(i*z) for all integers n.
 
  • #6
And you don't get the same equation for sin and cos. You are just being careless.
 
  • #7
Thx for checking it Dick, its good to know if I made a mistake :P. It felt funny that the same thing came out for sin that happened in cosine. I'll go back and redo the quadratic and see what went wrong.

Edit: Found problem, I divided by 2i instead of 2 when solving the quadratic.
 
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  • #8
Ok, redid the sin(z) and I got a different answer.
m = sin(z) = 1/2[e^iz - e^-iz]
2im = e^iz - e^-iz = e^-iz(e^2iz - 1)
2ime^iz = e^2iz - 1
e^2iz - 2ime^iz - 1 = 0
e^iz = [2im +/- (-4m^2 + 4)^(1/2)]/2
e^iz = I am +/- i(m^2 - 1)^(1/2)]
iz = Log[im +/- i(m^2 - 1)^(1/2)] mod2pi
z = -iLog[im +/- i(m^2 - 1)^(1/2)] mod2pi (Since its going in sin its periodic)

So, if m = 12i/5 then:

z = -iLog(-12/5 +/- i(-199/25 - 1)^(1/2))mod2pi
z = -iLog[-12/5 +/- -(198)^(1/2)/5]mod2pi

once you get these answers you can add or subtract steps of 2ipiK for all k in Z since log/the exponential are periodice by 2ipi.

Is there a way to easily simplify this Log statement so that the values are a little more concrete?

Using QuickMath Simplification... doesn't seem like you can take the log all that well. All it did was turn it into the form:

z = pi - iLog[1/5(12+(199)^(1/2)]
pi coming from Log(-1), taking 25 out of the sqroot and then taking 1/5 out of the expression. I gave my answer as above and then added steps of 2ipi for an infinite set of solutions. Thx for your help! If you find any mistakes reading this don't hestitate to remark!
 
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What is the meaning of surjectivity in the context of sin/cos over the complex plane?

Surjectivity refers to the property of a mathematical function where every element in the output set has at least one corresponding input element. In the case of sin and cos functions over the complex plane, it means that for every complex number in the output set (e.g. the unit circle), there is a corresponding complex number (or angle) in the input set (e.g. the real line).

Why is surjectivity important in the study of sin/cos over the complex plane?

Surjectivity is important because it allows us to fully understand the behavior and properties of the sin and cos functions over the complex plane. It also helps us to solve complex equations and analyze complex systems using these functions.

How is surjectivity related to the concepts of domain and range in mathematics?

Surjectivity is closely related to the concepts of domain and range. The domain of a function is the set of all possible input values, while the range is the set of all possible output values. For a function to be surjective, every element in the range must have at least one corresponding element in the domain.

Can the surjectivity of sin/cos over the complex plane be proven using mathematical proofs?

Yes, the surjectivity of sin and cos over the complex plane can be proven using mathematical proofs. One approach is to show that for every complex number z in the output set, there exists an angle θ in the input set such that sin(θ) = z and cos(θ) = z. This can be done using trigonometric identities and properties of complex numbers.

Are there any practical applications of understanding the surjectivity of sin/cos over the complex plane?

Yes, there are many practical applications of understanding the surjectivity of sin and cos over the complex plane. These functions are used in various fields such as physics, engineering, and computer graphics. Understanding their surjectivity allows us to model and analyze complex systems, solve equations, and create visual representations of data.

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