complex number relations

Question

I can structure a solution using geometry but I am struggling to find on using algebra.

In an Argand diagram, the complex numbers $z, w$ and $z+w$ are represented by the points $P, Q$ and $S$ respectively.

Show that: $$| z + w | ≥ | z | − | w |$$

Thanks for any help

Are you familiar with the triangule inequality $|u+v|\le |u|+|v|?$ — mfl, Sep 04 '17 at 16:23
So, are you looking for a proof of that? Since you have asked your question in terms of complex number I think this https://math.stackexchange.com/questions/582433/prove-the-triangle-inequality-involving-complex-numbers can be helpful. — mfl, Sep 04 '17 at 16:33

score 4 · Accepted Answer · answered Sep 04 '17 at 16:18

4

We have that

$$|z|=|z+w-w|\le |z+w|+|w|\implies |z|-|w|\le |z+w|.$$

answered Sep 04 '17 at 16:18

mfl

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This is a circular argument. – Johnathan Gross Sep 04 '17 at 16:21
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@JohnathanGross It isn't. It deduces the asked inequality from the triangular inequality. – mfl Sep 04 '17 at 16:22
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Two downvotes for a short and totally correct answer? Why? – José Carlos Santos Sep 04 '17 at 16:28
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It gets even worse. I just found out that someone suggested that this answer should be closed! – José Carlos Santos Sep 04 '17 at 16:40

Michael Rozenberg · Answer 2 · 2017-09-04T16:51:01.493

Let $z=x+yi$ and $w=a+bi$, where $\{x,y,a,b\}\subset\mathbb R$.

Thus, we need to prove that $$\sqrt{(x+a)^2+(y+b)^2}+\sqrt{a^2+b^2}\geq\sqrt{x^2+y^2}$$ or $$\sqrt{(a^2+b^2)((x+a)^2+(y+b)^2)}\geq-a^2-b^2-ax-by.$$ If $-a^2-b^2-ax-by<0$ then the inequality is obviously true, but for

$-a^2-b^2-ax-by\geq0$ it's enough to prove that $$(a^2+b^2)((x+a)^2+(y+b)^2)\geq(a^2+b^2+ax+by)^2,$$ which is C-S: $$(a^2+b^2)((x+a)^2+(y+b)^2)\geq(a(x+a)+b(y+b))^2.$$ Done!

If you don't like the C-S then for the proof of the last inequality we need to prove that $$(a^2+b^2)(a^2+b^2+2(ax+by)+x^2+y^2)\geq(a^2+b^2)^2+2(a^2+b^2)(ax+by)+(ax+by)^2$$ or $$(a^2+b^2)(x^2+y^2)\geq(ax+by)^2$$ or $$(ay-bx)^2\geq0.$$

@pythonstudent I don't understand you. I thought that you look for an algebraic proof. The proof by the inequality $|z+w|+|w|=|z+w|+|-w|\geq|z+w-w|=|z|$, it's a geometric proof because it's just the triangle inequality. — Michael Rozenberg, Sep 04 '17 at 16:56

score 0 · Answer 3 · answered Sep 04 '17 at 17:12

Other answers are perfectly correct, here is an answer using cosine rule:

$$\begin{align}|z-w|^2 &=(z-w)\overline{(z-w)} \\ &=|z|^2 +|w|^2 - z\overline{w} - \overline{z}w \\ &= |z|^2 + |w|^2 - (z\overline{w} + \overline{z\overline{w}} )\\ &= |z|^2 + |w|^2 - 2\Re{(z\overline{w})} \\ &\ge |z|^2 - |w|^2 -2|z\overline{w}| \\ &\ge (|z| - |w|)^2 \end{align}$$

$|\Re(z)| \le|z|$ as $\Re^2(z) + \Im^2(z) =|z|^2$

Now since $$|z|^2 +|w|^2 + z\overline{w} + \overline{z}w \ge |z|^2 +|w|^2 - z\overline{w} - \overline{z}w$$ $$\implies |z+w|^2 \ge|z-w|^2$$

complex number relations

3 Answers3