It's rather easy to show that:

$$\begin{aligned} \left|z\right|^2 = z \cdot \overline z \,\,\,\,\,\text{ where } z = x + iy \end{aligned}$$

How about:

$$\begin{aligned} \left | z + 1 \right |^2 = (z + 1) \cdot (\,\overline z + 1) \end{aligned}$$

?

As a challenge, consider:

$$\begin{aligned} Re(e^{\frac{z}{z+1}}) = \left( e^{\frac{\left|z\right|^2+x}{\left|z+1\right|^2}}\right) \cdot \cos\left(\frac{y}{\left| z+1 \right| ^2}\right) \end{aligned}$$

where the left hand side refers to the real part of $\displaystyle e^{\frac{z}{z+1}}$. Does the equation hold ? Note in particular the right hand side is a real number, but in terms of numbers in the complex plane !

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 11:44 PM