Is z/(1+|z|) continuous on the complex plane

The simplest way to prove continuity is to note that z is continuous so |z| is and therefore 1 + |z| is and since 1 + |z| > 0 for all z, 1/(1 + |z|) is also continuous so that z/(1 + |z|) is continuous for all z.

But if e-d you must, then..

In order to prove continuity at z0 in C, choose d = e/(1 + 2|z0|))

If |z - z0| < d, we have

|z/(1 + |z|) - z0/(1 + |z0|)| = | z(1 + |z0|) - z0(1 + |z|) |/ |(1 + |z|)(1 + |z0|)|

<= |z + z|z0| - z0 - z0|z| |

<= |z - z0 + z|z0| - z0|z| |

<= |z - z0| + |z|z0| - z0|z| |

= |z - z0| + | (z - z0)|z0| + z0|z0| - z0|z| |

= |z - z0| + | (z - z0)|z0| + z0(|z0| - |z|) |

<= |z - z0| + |z - z0||z0| + |z0|||z0| - |z||

= |z - z0| + |z - z0||z0| + |z0|||z| - |z0||

= |z - z0| + |z0|( |z - z0| + ||z| - |z0|| )

<= |z - z0| + |z0|( |z - z0| + |z - z0| )

= |z - z0| + 2|z0|(|z - z0|)

= |z - z0|(1 + 2|z0|)

< d(1 + 2|z0|)

= [ e/(1 + 2|z0|)) ] (1 + 2|z0|)

= e

Note: I made use of the following incarnations of the Triangle Inequality in the above proof:

|z + w| <= |z| + |w|
|z - w| <= |z| + |w|
||z| - |w|| <= |z - w|

Sure, why not?

How is the continuity of z / (1 + |z|) any different than the continuity of z itself? If you think it is discontinuous, let us know why and we can discuss it.