MHB Proving Modulus Inequality for Complex Numbers

[cbarker1]

Dear Everyone,

I am currently in an Introduction to Complex Analysis; I have a question:

Use established properties of moduli to show that when $\left|{z_3}\right|\ne\left|{z_4}\right|$:
$\frac{\Re{({z}_{1}+{z}_{2})}}{\left|{z}_{3}+{z}_{4}\right|}\le\frac{\left| {z}_{1} \right|+\left| {z}_{2} \right|}{\left| \left| {z}_{3} \right|-\left| {z}_{4} \right| \right|}$

My work:
Let ${z}_{1},{z}_{2}\in\Bbb{C}$ and ${z}_{3}, {z}_{4}$ be non-zero complex numbers.

I am stuck.

Thanks
CBarker1

 

[I like Serena]

Dear Everyone,

I am currently in an Introduction to Complex Analysis; I have a question:

Use established properties of moduli to show that when $\left|{z_3}\right|\ne\left|{z_4}\right|$:
$\frac{\Re{({z}_{1}+{z}_{2})}}{\left|{z}_{3}+{z}_{4}\right|}\le\frac{\left| {z}_{1} \right|+\left| {z}_{2} \right|}{\left| \left| {z}_{3} \right|-\left| {z}_{4} \right| \right|}$

My work:
Let ${z}_{1},{z}_{2}\in\Bbb{C}$ and ${z}_{3}, {z}_{4}$ be non-zero complex numbers.

I am stuck.

Thanks
CBarker1

Hi Cbarker1,

Have you considered the triangle inequalities?
With complex numbers they are:
$$\begin{array}{} |z_1| + |z_2| \ge |z_1+z_2| & \text{Sum of 2 sides of a triangle}\ge\text{3rd side}\\
\Big||z_1|-|z_2|\Big| \le |z_1+z_2| & \text{Difference of 2 sides of a triangle}\le\text{3rd side}\\
\operatorname{Re} z \le |z| & \text{Rectangular side in a right triangle}\le \text{hypotenuse}
\end{array}$$

How far can you get with them?

 

[cbarker1]

Hi Cbarker1,

Have you considered the triangle inequalities?
With complex numbers they are:
$$\begin{array}{} |z_1| + |z_2| \ge |z_1+z_2| & \text{Sum of 2 sides of a triangle}\ge\text{3rd side}\\
\Big||z_1|-|z_2|\Big| \le |z_1+z_2| & \text{Difference of 2 sides of a triangle}\le\text{3rd side}\\
\operatorname{Re} z \le |z| & \text{Rectangular side in a right triangle}\le \text{hypotenuse}
\end{array}$$

How far can you get with them?

I believe the solution is this:
$\frac{\Re{({z}_{1}+{z}_{2}})}{\left|{z}_{3}+{z}_{4}\right|}\le\frac{\left| {z}_{1}\right|+\left|{z}_{2} \right|}{\left|\left| {z}_{3} \right|-\left| {z}_{4} \right|\right|}$. By the triangle inequality and reverse triangle inequality, the inequality yields
$\le\frac{\left| {z}_{1}+{z}_{2} \right|}{\left|{z}_{3}+{z}_{4}\right|}$
By the $\operatorname{Re}{z}\le\left| z \right|$, then the inequality yields
$\le\frac{\operatorname{Re}{({z}_{1}+{z}_{2}})}{\left|{z}_{3}+{z}_{4}\right|}$ QED

 

[Euge]

I believe the solution is this:
$\frac{\Re{({z}_{1}+{z}_{2}})}{\left|{z}_{3}+{z}_{4}\right|}\le\frac{\left| {z}_{1}\right|+\left|{z}_{2} \right|}{\left|\left| {z}_{3} \right|-\left| {z}_{4} \right|\right|}$. By the triangle inequality and reverse triangle inequality, the inequality yields
$\le\frac{\left| {z}_{1}+{z}_{2} \right|}{\left|{z}_{3}+{z}_{4}\right|}$
By the $\operatorname{Re}{z}\le\left| z \right|$, then the inequality yields
$\le\frac{\operatorname{Re}{({z}_{1}+{z}_{2}})}{\left|{z}_{3}+{z}_{4}\right|}$ QED

You seem to be going backwards, rather than forwards, in your argument. The goal is to show $\displaystyle \frac{\Re(z_1 + z_2)}{|z_3 + z_4|} \le \frac{|z_1| + |z_2|}{||z_3| - |z_4||}$. Since $\Re z \le |z|$ for all complex numbers $z$, $\displaystyle \frac{\Re(z_1 + z_2)}{|z_3 + z_4|} \le \frac{|z_1 + z_2|}{|z_3 + z_4|}$. The triangle and reverse triangle inequalities yields $|z_1 + z_2| \le |z_1| + |z_2|$ and $|z_3 + z_4| \ge ||z_3| - |z_4||$; hence $\displaystyle \frac{|z_1 + z_2|}{|z_3 + z_4|} \le \frac{|z_1| + |z_2|}{||z_3| - |z_4||}$, as desired.