Wednesday, November 06, 2013
\(\left| z \right|^2 = z \cdot \overline z\)
It's rather easy to show that:
  | \[ \begin{aligned} \left|z\right|^2 = z \cdot \overline z \,\,\,\,\,\text{ where } z = x + iy \end{aligned} \] |
  | \[ \begin{aligned} \left | z + 1 \right |^2 = (z + 1) \cdot (\,\overline z + 1) \end{aligned} \] |
As a challenge, consider:
  | \[ \begin{aligned} Re(e^{\frac{z}{z+1}}) = \left( e^{\frac{\left|z\right|^2+x}{\left|z+1\right|^2}}\right) \cdot \cos\left(\frac{y}{\left| z+1 \right| ^2}\right) \end{aligned} \] |
More at Analysis of a Complex Kind.