# Absolute Value of Components of Complex Number no greater than Root 2 of Modulus

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## Theorem

Let $z = x + i y \in \C$ be a complex number.

Then:

$\size x + \size y \le \sqrt 2 \cmod z$

where:

$\size x$ and $\size y$ denote the absolute value of $x$ and $y$
$\cmod z$ denotes the complex modulus of $z$.

## Proof

Let $z = x + i y \in \C$ be an arbitrary complex number.

Aiming for a contradiction, suppose the contrary:

 $\ds \size x + \size y$ $>$ $\ds \sqrt 2 \cmod z$ $\ds \leadsto \ \$ $\ds \paren {\size x + \size y}^2$ $>$ $\ds 2 \cmod z^2$ squaring both sides $\ds \leadsto \ \$ $\ds \size x^2 + 2 \size x \, \size y + \size y^2$ $>$ $\ds 2 \cmod z^2$ multiplying out $\ds \leadsto \ \$ $\ds x^2 + 2 \size x \, \size y + y^2$ $>$ $\ds 2 \cmod z^2$ Definition of Absolute Value $\ds \leadsto \ \$ $\ds x^2 + 2 \size x \, \size y + y^2$ $>$ $\ds 2 \paren {x^2 + y^2}$ Definition of Complex Modulus $\ds \leadsto \ \$ $\ds 2 \size x \, \size y$ $>$ $\ds x^2 + y^2$ $\ds \leadsto \ \$ $\ds 2 \size x \, \size y$ $>$ $\ds \size x^2 + \size y^2$ Definition of Absolute Value $\ds \leadsto \ \$ $\ds \size x^2 - 2 \size x \, \size y + \size y^2$ $<$ $\ds 0$ rearranging $\ds \leadsto \ \$ $\ds \paren {\size x - \size y}^2$ $<$ $\ds 0$ factoring

But as $\size x$ and $\size y$ are both real this cannot happen.

Thus our initial assumption $\size x + \size y > \sqrt 2 \cmod z$ is false.

Hence the result.

$\blacksquare$