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Someone could help to prove the following inequality of modulus of complex numbers:

If $a\in\mathbb{C}$ then $$|a|\leq|a+z| \qquad \forall z\in\mathbb{C}$$

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3 Answers 3

It will be hard to prove. Try $a=1, z=-1$ where it is false.

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It isn't true.

$$a=1+i,\;z=-1-i$$

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and if $z\neq -a$? –  Roiner Segura Cubero Oct 1 '12 at 23:40
    
@Andres Still not I'm afraid - example: $a=i,\;z=-i/2$ –  user39572 Oct 1 '12 at 23:47
    
Julien and this inequality is true? –  Roiner Segura Cubero Oct 1 '12 at 23:56
    
Julien and this inequality is true?\\\\ $$|a_n|\leq\left|a_n+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}+...+\frac{a_{1}}{z^{n‌​-1}}+\frac{a_{0}}{z^{n}}\right|$$ with $z\neq 0$, $a_n\neq 0$ and $a_k\in \mathbb{C}$ for $k=0, 1,...,n$ –  Roiner Segura Cubero Oct 2 '12 at 0:04
    
Now if Julien!! I'm trying to make a proof of the fundamental theorem of algebra using the polynomial $ P (z) = a_ {0} + a_ {1} z + ... + a_ {n} z ^ {n} $ with $ a_ {n } \neq 0 $. and let me know if the above inequality is true –  Roiner Segura Cubero Oct 2 '12 at 0:11

This is false(!) Try $a = 1$, $z$ any negative number in $[-2, 0]$.

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and if $|z|\rightarrow\infty$? –  Roiner Segura Cubero Oct 2 '12 at 1:30
    
For larger $z$, then your inequality can hold. But not for all $z$. But you said "$\forall z \in \mathbb{C}$". –  mike4ty4 Oct 2 '12 at 1:43

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