User:Anghel/Sandbox

Theorem
Let $\C$ denote the complex plane.

Let $R^2$ denote the Euclidean plane.

Then the function $\phi : \R^2 \to \C$ defined by:


 * $\map \phi { x, y } = x + i y$

is a homeomorphism between $R^2$ and $\C$.

Proof
Define $\phi^{-1} : \C \to \R^2$ by:


 * $\map { \phi^{-1} }{ z } = \paren { \map \Re z, \map \Im z }$

Then $\phi^{-1}$ is the inverse of $\phi$, as:


 * $\map { \phi }{ \map { \phi^{-1} }{ z } } = \map \Re z + i \map \Im z = z $


 * $\map { \phi^{-1} }{ \map { \phi }{ x, y } } = \paren { \map \Re { x + i y } , \map \Im { x + i y } } = \paren { x , y }$

By definition of bijection, it follows that $\phi$ is bijective.

We show continuity of $\phi$.

Complex Plane is Metric Space shows that for $\map \phi { x_1, y_1 } = x_1 + i y_1 , \map \phi { x_2 , y_" } = x_2 + i y_2$, we have


 * $ \cmod { \map \phi { x_2, y_2 } - \map \phi { x_1 , y_1 } } = \sqrt { \paren { x_2 - x_1 }^2 + \paren { y_2 - y_1 }^2 }$

which is equal to the distance between $\paren { x_1, x_2 }$ and $\paren { y_1 , y_2 }$ in the euclidean metric on the Euclidean plane.

Let $\epsilon \in \R_{ >0 }$.

With $\delta = \epsilon$, we have:


 * $\sqrt { \paren { x_2 - x_1 }^2 + \paren { y_2 - y_1 }^2 } < \delta \implies \cmod { z_2 - z_1 } < \epsilon$

which shows the continuity of $\phi$.

The continuity of $\phi^{-1}$ is proven the same way.

RemCategory:Complex Analysis