# A problem in comlex analysis

• Oct 23rd 2008, 04:34 AM
Itzic2006
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.
• Oct 23rd 2008, 06:20 AM
Itzic2006
A little fix...
The condition that needs to be fulfilled is : |xz1 + yz2| >= d|x+y|, and not like it was written in the opening post.
Sorry..
• Oct 24th 2008, 05:26 AM
Opalg
Define $f:\mathbb{R}^2\to\mathbb{R}$ by $f(x,y) = |xz_1+yz_2|$. Notice that $f(x,y)\ne0$ whenever $(x,y)\ne(0,0)$ (because $xz_1+yz_2=0\Rightarrow z_1/z_2$ is real).

The set $S = \{(x,y)\in\mathbb{R}^2:|x|+|y| = 1\}$ is closed and bounded, therefore compact, and so the function f(x,y) attains its minimum value d (> 0) on this set. Therefore $|xz_1+yz_2| \geqslant d(|x|+|y|) \geqslant d|x+y|$ on S and hence (by taking scalar multiples) on the whole of $\mathbb{R}^2$.
• Oct 24th 2008, 05:35 AM
Laurent
Quote:

Originally Posted by Itzic2006
The condition that needs to be fulfilled is : |xz1 + yz2| >= d|x+y|, and not like it was written in the opening post.
Sorry..

First there are various ways to simplify the problem:
- by factoring by $z_1$, it suffices to prove that, if $z$ is a complex non-real number, there is $d>0$ such that, for every $x,y\in\mathbb{R}$, $|x+yz|\geq d|x+y|$.
- because $\frac{x+yz}{x+y}=\frac{x}{x+y}+\frac{y}{x+y}z$, it suffices to prove that, if $z$ is a complex non-real number, there is $d>0$ such that, for every $u\in\mathbb{R}$, $|(1-u)+uz|>d$. (the $u$ stands for $\frac{y}{x+y}$ in the previous formulation).

This is the same as proving that $|(1-u)+uz|^2>d^2$. However, $|(1-u)+uz)|^2=(1-u)^2 + 2u(1-u){\rm Re}(z)+u^2|z|^2$ is just a second-order polynomial in the (real) variable $u$, which is always positive because $(1-u)+uz=0$ would imply that $z$ is real. As a consequence, it is bounded from below by a positive constant. If you procede to the explicit computation of the minimum of this polynomial, you get that it is $\frac{b^2}{(a-1)^2+b^2}$ if $z=a+ib$, $a,b\in\mathbb{R}$. This gives the minimum in the original question if you replace $z=\frac{z_2}{z_1}$ and take a square root: the minimum is $\frac{{|\rm Im}\left(\frac{z_2}{z_1}\right)|}{\left|\frac{z_2} {z_1}-1\right|}=\left|\sin{\rm
Arg}\left(\frac{z_2}{z_1}-1\right)\right|$
(sinus of the argument of...).

You can prove the property of the last paragraph geometrically: the points $(1-u)+uz$, $u\in\mathbb{R}$ lie on a line that goes through $1$ and $z$ (for $u=1$), and we want the distance of this line to $0$. It is achieved by the orthogonal projection of $0$ on the line, and trigonometry gives its expression as a sinus.

If you're not looking for explicit $d$, you could also have said (this is not much different from what I wrote before): we easily see that the application $N:(x,y)\mapsto |z_1 x+z_2 y|$ is a norm on $\mathbb{R}^2$ ( $N(x,y)=0$ implies $x=y=0$ because $(z_1,z_2)$ is an $\mathbb{R}-$basis of $\mathbb{C}$), hence $\frac{N(x,y)}{|x+y|}=N\left(\frac{x}{x+y},\frac{y} {x+y}\right)$ is greater that the distance (for the norm $N$) between $(0,0)$ and the line $\{(u,v)|u+v=1\}$. This line is a closed set not containing 0, so that this distance is positive, and we are done.
• Oct 24th 2008, 05:42 AM
Laurent
Quote:

Originally Posted by Opalg
Define $f:\mathbb{R}^2\to\mathbb{R}$ by $f(x,y) = |xz_1+yz_2|$. Notice that $f(x,y)\ne0$ whenever $(x,y)\ne(0,0)$ (because $xz_1+yz_2=0\Rightarrow z_1/z_2$ is real).

The set $S = \{(x,y)\in\mathbb{R}^2:|x|+|y| = 1\}$ is closed and bounded, therefore compact, and so the function f(x,y) attains its minimum value d (> 0) on this set. Therefore $|xz_1+yz_2| \geqslant d(|x|+|y|) \geqslant d|x+y|$ on S and hence (by taking scalar multiples) on the whole of $\mathbb{R}^2$.

Nice solution, I like it.
My post is still interesting if the exact minimum is needed.
• Oct 24th 2008, 07:18 AM
Itzic2006
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 !
• Oct 24th 2008, 07:50 AM
Laurent
Quote:

Originally Posted by Itzic2006
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 !!

I felt like what I was writing could be unclear. Let's do it again without too much words, because basically it is very simple:
First, since $z_1\neq 0$, I can write $\frac{|xz_1+yz_2|}{|x+y|}=|z_1|\left|\frac{x}{x+y} +\frac{y}{x+y}\frac{z_2}{z_1}\right|.$
From there, to ease notations, I let $z=z_2/z_1$ and $u=\frac{y}{x+y}$ (so that $\frac{x}{x+y}=1-u$). This gives: $\frac{|xz_1+yz_2|}{|x+y|}=|z_1||(1-u)+uz|$ and $u$ is a real number.
If $z=a+ib$, we have $(1-u)+uz=(1-u)+ua +i ub$, hence $|(1-u)+uz|^2=(1-u+ua)^2+(ub)^2$.
Now, expand this: this is a polynomial of degree two in the variable $u$. You can compute its minimum value by studying the variations of this function. This gives you $\frac{b^2}{(a-1)^2+b^2}$. Finally, you have:
$\frac{|xz_1+yz_2|}{|x+y|}\geq|z_1|\sqrt{\frac{b^2} {(a-1)^2+b^2}}=|z_1|\frac{|b|}{\sqrt{(a-1)^2+b^2}}$ and you are done (because $b\neq 0$ and $z_1\neq 0$), with an explicit (and optimal) $d$. This is not important, but the ratio can be seen as a sinus because it is the imaginary part of $z$ (or $z-1$) divided by modulus of $z-1$.
(And I notice that I forgot the $|z_1|$ in my previous post...)
• Oct 24th 2008, 08:36 AM
Opalg
Quote:

Originally Posted by Itzic2006
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 !

Yes, that was exactly what I meant. (Clapping)