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I Circular Functions

  1. Aug 10, 2017 #1
    Does a circular function with complex variable represent a three-dimensional graph?

    For example cosiz
  2. jcsd
  3. Aug 10, 2017 #2


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    The definition of [itex]\cos(w) [/itex] with w complex is [itex] \cos(w)=\frac{e^{i\cdot w}+e^{-i\cdot w}}{2}[/itex]. Substitute [itex] w=i\cdot z[/itex] and you get [itex] \cos(i\cdot z)=\frac{e^{i\cdot (i\cdot z)}+e^{-i\cdot (i\cdot z)}}{2}=\frac{e^{-z}+e^{z}}{2}[/itex]. Looks familiar?
  4. Aug 11, 2017 #3
    I have read that 'i' represent the rotation of a sphere. And I have understood that similar to a two dimensional function which forms a quadratic equation, the rotation of sphere along its three dimensional axis will form a cubic equation whose roots contain complex numbers. And my question is that does a hyperbolic function that contains complex variable represent a 3-dimensional geometry in the same way a circular function represent a 2-dimensional geometry?
  5. Aug 12, 2017 #4


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    Well, no. It represents a 90° rotation of the coordinate system.
    I have absolutely no idea of what this means.
    As I demonstrated above, the hyperbolic and circular functions are just a 90° rotation away from each other. You can combine them in different fashions, for example (assuming z=x+iy): [itex] \vert \cos(z) \vert ^{2}=\sinh(y)^{2}+\cos(x)^{2}=\cosh(y)^{2}-\sin(x)^{2}=\frac{1}{2}(\cosh(2y)+\cos(2x))[/itex]
  6. Aug 12, 2017 #5
    Can you clarify me around which axis the coordinate system is rotated 90°? Is the rotation happening alongside a different dimension than the xyz dimension?
  7. Aug 13, 2017 #6


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    Forget the "xyz dimension". The complex plane is a plane, with the real axis corresponding to the "x-axis" and the imaginary axis corresponding to the "y-axis". As you know, it is no problem to rotate the real "xy-plane" 90° without messing around with any third axis. You can describe it as x→y; y→-x or use a rotation matrix: [itex]
    0 & 1 \\
    -1 & 0 \\
    Now: the complex plane has its own version of these rules, making rotations very simple. A rotation with an angle of φ corresponds to a multiplication with [itex] e^{i\varphi}[/itex]. Thus, rotating 90° (which in math term is π/2) means multiplying with [itex] e^{i\frac{\pi}{2}}[/itex]. But as [itex] e^{i\frac{\pi}{2}}=i[/itex], multiplying with i is equivalent with a 90° rotation.
  8. Aug 13, 2017 #7
    Any ##f:\mathbb{C} \to \mathbb{C}## represents a 2d vector to another 2d vector, so the graph of any such function would be represented by four dimensions.
    Last edited: Aug 13, 2017
  9. Aug 13, 2017 #8


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    z = x+iy has two real dimensions (x,y) and cos( iz ) = u(iz) + iv(iz) also has two real dimensions (u,v). So it can be considered a two closely related 3-dimensional graphs. One is the graph of u as a function of (x,y) and the other is a graph of v as a function of (x,y).

    In studying complex analysis, you will learn that since cos( iz ) is a holomorphic function, u and v are called harmonic functions and are related to each other by the Cauchy-Riemann equations: ∂u/∂x = ∂v/∂y; ∂u/∂y = -∂v/∂x.
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