Equation of Unit Circle in Complex Plane/Proof 2

Theorem
Consider the unit circle $C$ whose center is at $\left({0, 0}\right)$ on the complex plane.

The equation of $C$ is given by:
 * $\left\vert{z}\right\vert = 1$

where $\left\vert{z}\right\vert$ denotes the complex modulus of $z$.

Proof
Let $C$ be the set of points on the unit circle defined above.

Let $P$ be the set:
 * $\left\{{z \in \C: \left\vert{z}\right\vert = 1}\right\}$

Let $z = x + i y \in C$.

$z$ can be expressed in polar form as:
 * $z = \left\langle{r, \theta}\right\rangle$

where:
 * $x = r \cos \theta$
 * $y = r \sin \theta$

By definition of unit circle, $z$ is $1$ unit away from $\left({0, 0}\right)$.

By Pythagoras's Theorem:
 * $\sqrt {\left({r \cos \theta}\right)^2 + \left({r \sin \theta}\right)^2} = 1$

from which:
 * $\sqrt {x^2 + y^2} = 1$

By definition of complex modulus:
 * $\left\vert{z}\right\vert = 1$

and so $z \in P$.

Thus $C \subseteq P$.

Let $z \in P$.

Then by definition $\left\vert{z}\right\vert = 1$.

By Modulus of Complex Number equals its Distance from Origin, $z$ is $1$ unit away from $\left({0, 0}\right)$.

By definition of $C$, it follows that $z \in C$.

Thus $P \subseteq C$.

The result follows by definition of set equality.