MHB Helpful Tips on Solving Complex Equations

[fabiancillo]

Hi! I have problems with this demonstration

Let $z= x+iy , x,y \in \mathbb{R} $ then $|x|, |y| \leq{|z|} \leq{\sqrt[ ]{2}} $ , $max \{ |x|, |y| \} $

 

[Greg]

Hello cristianoceli and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[fabiancillo]

Hello cristianoceli and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

Si! my English is not very good. I'm looking for a suggestion of how to start

Sorry for not upload anything but I do not know how to start

Thanks

 

[MarkFL]

Should the problem actually read:

Let $z= x+iy,\,x,y\in\mathbb{R}$ then $|x|,\,|y|\leq|z|\leq\sqrt{2}\cdot\max(|x|,|y|)$ ?

 

[fabiancillo]

Should the problem actually read:

Let $z= x+iy,\,x,y\in\mathbb{R}$ then $|x|,\,|y|\leq|z|\leq\sqrt{2}\cdot\max(|x|,|y|)$ ?

Why $\cdot$ ?

 

[MarkFL]

Why $\cdot$ ?

It makes more sense to me than a comma...it implies multiplication. :)

 

[Prove It]

The left hand inequality is easy if you remember the definition of a modulus for real and complex numbers:

Real numbers: $\displaystyle \begin{align*} \left| x \right| = \sqrt{x^2} \end{align*}$.

Complex numbers: $\displaystyle \begin{align*} \left| z \right| = \sqrt{x^2 + y^2} \end{align*}$.