A problem in comlex analysis

**Hi everyone,**

I have a rather annoying question in compex analysis. The question, I believe, should be solved without any use of advanced

compex analysis theorems or lemas, because it's in the first chapter in my book, so basic "calculus tricks" should suffice.

And for the problem : Let Z1,Z2 be two comlex numbers which fulfills : Z1/Z2 is not a real number (meaning the two vectors are not proportionate to one another), and Z2 is not 0 of course.

Prove that there's a real positive number - d>0 - that for each two real numbers x,y the following term will be fulfilled :

|xz1 + yz2| >= |x+y|

I'm really frustrated with that question, coming very close to solve at any time I take a different approach, but not there quite yet !

Any help/idea would be willingly recieved and appreciated.

Thanks,

Itzic T.

A gratitude and a little wonder :)

First of all, let me thank you both Laurent and Opalg for your help.

Laurent, It was quite difficult for me to get through all your proof. It seems like my level in math is no where near yours, and I'm not used to read math in English so this made my understanding of you fabulous post only partial. But thank you very much anyway !!

Opalg, I did undserstand what you wrote. The only part I'm not quite sure what you meant was the end.

By writing : "and hence (by taking scalar multiples) on the whole of http://www.mathhelpforum.com/math-he...f52dce48-1.gif "

did you mean that for every (x,y) != (0,0) you can divide the term

|XZ1+yZ2| by |x|+|y| and therfore to get that x' = x/(|x|+|y|) and

y' = y/(|x|+|y|) will obtain : |x'| +|y'| = 1 and (x',y') belong to S (and from there the original inequality derives ?).

If it was not what you meant I'll be glad to get you short clarification on another post. Thank you cery much !