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

This entry was written by Nalin Pithwa, posted on February 18, 2015 at 7:07 pm, filed under IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) and tagged . Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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