Complex numbers

Question: Prove that for any complex number z:

$|z+1| \geq \frac {1}{\sqrt {2}}$ or $|z^{2}+1| \geq 1$

Solution: Suppose by way of contradiction that

$|1+z| < \frac {1}{\sqrt {2}}$ and $|1+z^{2}| <1$

Setting $z=a+ib$ with $a,b \in R$ yields $z^{2}=a^{2}-b^{2}+i2ab$ .

We obtain

$(1+a^{2}-b^{2})^{2}+4a^{2}b^{2}<1$ and $(1+a)^{2}+b^{2}<1/2$

and consequently,

$(a^{2}+b^{2})^{2}+2(a^{2}-b^{2})<0$ and $2(a^{2}+b^{2})+4a+1<0$ .

Summing these inequalities implies $(a^{2}+b^{2})^{2}+(2a+1)^{2}>0$ which is a contradiction.

More later,

Nalin Pithwa

Mathematics Hothouse