Can a Circular Function with Complex Variable Represent a 3D Graph?

[Leo Authersh]

Does a circular function with complex variable represent a three-dimensional graph?

For example cosiz

 

[Svein]

The definition of $\cos(w)$ with w complex is $\cos(w)=\frac{e^{i\cdot w}+e^{-i\cdot w}}{2}$. Substitute $w=i\cdot z$ and you get $\cos(i\cdot z)=\frac{e^{i\cdot (i\cdot z)}+e^{-i\cdot (i\cdot z)}}{2}=\frac{e^{-z}+e^{z}}{2}$. Looks familiar?

 

[Leo Authersh]

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?

 

[Svein]

I have read that 'i' represent the rotation of a sphere.

Well, no. It represents a 90° rotation of the coordinate system.

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.

I have absolutely no idea of what this means.

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?

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): $\vert \cos(z) \vert ^{2}=\sinh(y)^{2}+\cos(x)^{2}=\cosh(y)^{2}-\sin(x)^{2}=\frac{1}{2}(\cosh(2y)+\cos(2x))$

 

[Leo Authersh]

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): $\vert \cos(z) \vert ^{2}=\sinh(y)^{2}+\cos(x)^{2}=\cosh(y)^{2}-\sin(x)^{2}=\frac{1}{2}(\cosh(2y)+\cos(2x))$

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?

 

[Svein]

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?

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: $\begin{pmatrix}<br /> 0 & 1 \\<br /> -1 & 0 \\<br /> \end{pmatrix}$.
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 $e^{i\varphi}$. Thus, rotating 90° (which in math term is π/2) means multiplying with $e^{i\frac{\pi}{2}}$. But as $e^{i\frac{\pi}{2}}=i$, multiplying with i is equivalent with a 90° rotation.

 

[Someone2841]

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.

 

[FactChecker]

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.