Equation of Unit Circle in Complex Plane

From ProofWiki
Jump to navigation Jump to search

Theorem

Consider the unit circle $C$ whose center is at $\tuple {0, 0}$ on the complex plane.

Its equation is given by:

$\cmod z = 1$

where $\cmod z$ denotes the complex modulus of $z$.


Corollary 1

The equation of $C$ can be given by:

$z \overline z = 1$

where $\overline z$ denotes the complex conjugate of $z$.


Corollary 2

The equation of $C$ can be given by:

$\overline z = \dfrac 1 z$

where $\overline z$ denotes the complex conjugate of $z$.


Proof 1

From Equation of Unit Circle, the unit circle whose center is at the origin of the Cartesian $xy$ coordinate plane has the equation:

$x^2 + y^2 = 1$

Identifying the Cartesian $xy$ coordinate plane with the complex plane:



Proof 2

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

Let $P$ be the set:

$\set {z \in \C: \cmod z = 1}$

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

$z$ can be expressed in polar form as:

$z = \polar {r, \theta}$

where:

$x = r \cos \theta$
$y = r \sin \theta$

By definition of unit circle, $z$ is $1$ unit away from $\tuple {0, 0}$.

By Pythagoras's Theorem:

$\sqrt {\paren {r \cos \theta}^2 + \paren {r \sin \theta}^2} = 1$

from which:

$\sqrt {x^2 + y^2} = 1$

By definition of complex modulus:

$\cmod z = 1$

and so $z \in P$.

Thus $C \subseteq P$.


Let $z \in P$.

Then by definition $\cmod z = 1$.

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

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

Thus $P \subseteq C$.

The result follows by definition of set equality.

$\blacksquare$


Sources