# Complex Arithmetic/Examples/(Modulus of (z 1^2 + conj z 2^2))^2 + (Modulus of (conj z 3^2 - z 2^2))^2

## Example of Complex Arithmetic

Let $z_1 = 1 - i$, $z_2 = -2 + 4 i$ and $z_3 = \sqrt 3 - 2 i$.

Then:

$\cmod { {z_1}^2 + \overline {z_2}^2}^2 + \cmod {\overline {z_3}^2 - {z_2}^2}^2 = 765 + 128 \sqrt 3$

## Proof

 $\ds \cmod { {z_1}^2 + \overline {z_2}^2}^2 + \cmod {\overline {z_3}^2 - {z_2}^2}^2$ $=$ $\ds \cmod {\paren {1 - i}^2 + \overline {\paren {-2 + 4 i} }^2}^2 + \cmod {\overline {\paren {\sqrt 3 - 2 i} }^2 - \paren {-2 + 4 i}^2}^2$ $\ds$ $=$ $\ds \cmod {\paren {1 - i}^2 + \paren {-2 - 4 i}^2}^2 + \cmod {\paren {\sqrt 3 + 2 i}^2 - \paren {-2 + 4 i}^2}^2$ $\ds$ $=$ $\ds \cmod {\paren {1 - 2 i + i^2} + \paren {4 + 16 i + 16 i^2} }^2 + \cmod {\paren {3 + 4 \sqrt 3 i + 4 i^2} - \paren {4 - 16 i + 16 i^2} }^2$ $\ds$ $=$ $\ds \cmod {-12 + 14 i}^2 + \cmod {11 + \paren {16 + 4 \sqrt 3} i}^2$ $\ds$ $=$ $\ds \paren {144 + 196} + \paren {121 + \paren {16 + 4 \sqrt 3}^2}$ $\ds$ $=$ $\ds 461 + 256 + 48 + 128 \sqrt 3$ $\ds$ $=$ $\ds 765 + 128 \sqrt 3$

$\blacksquare$