inequality of modulus of a complex number

Question

To prove $|a+ib| \leq |a|+|b|$,

we know that $|a+ib|$ of a complex number is $\sqrt{a^2+b^2}$,

as $a$ and $b$ both are real numbers $a^2\geq0, b^2\geq0$, therefore $$\sqrt{a^2+b^2} \leq \sqrt{a^2}+\sqrt{b^2}$$ because $\sqrt{a+b} \leq \sqrt{a}+\sqrt{b} $ as a,b are non negative, and therefore $$|a+ib|= \sqrt{a^2+b^2} \leq \sqrt{a^2}+\sqrt{b^2} = |a|+|b|$$ is this argument correct? if not why? and kindly tell me whether there are any other ways to prove this thing.

$a$ is a real numbers then $a^2>0$? Isn't $0$ a real number? — jjagmath, Mar 25 '23 at 03:48
@PraveenKumaranP "because $\sqrt{a+b} \leq \sqrt{a}+\sqrt{b}$" $;$ You'll have to prove that, not just state it. — dxiv, Mar 25 '23 at 06:10
ok, let $a>0 ,b>0,$ wkt, $a^2 \leq b^2$ iff $a \leq b$ , using that $a+b \leq a+b+2\sqrt{a}\sqrt{b}$ , but since $a,b>0$, $2\sqrt{a}\sqrt{b} >0$, and hence the inequality. — Praveen Kumaran P, Mar 25 '23 at 06:31
@PraveenKumaranP That works, but you could prove $\sqrt{a^2+b^2} \le \sqrt{a^2} + \sqrt{b^2}$ directly the same way. That's what this answer does. — dxiv, Mar 25 '23 at 17:12

score 1 · Answer 1 · answered Mar 25 '23 at 03:35

1

You can apply the triangle inequality where $|i| = 1$:

\begin{align*} |a + ib| \leq |a| + |ib| = |a| + |i||b| = |a| + |b| \end{align*}

More generally, one has that: \begin{align*} |a + e^{i\theta}b| \leq |a| + |e^{i\theta}b| = |a| + |e^{i\theta}||b| = |a| + |b| \end{align*}

Hopefully this helps!

answered Mar 25 '23 at 03:35

Átila Correia

18,937

its cool but we know the triangle inequality for the reals, but how can use for an expression which includes i, is there any results or statements, if yes kindly help me – Praveen Kumaran P Mar 25 '23 at 03:39
@PraveenKumaranP You are right that it's not immediate that the triangle inequality holds for complex numbers. This is worth proving in its own right, and you can find discussions on this on MSE (for example https://math.stackexchange.com/questions/582433/prove-the-triangle-inequality-involving-complex-numbers) – pancini Mar 25 '23 at 03:42

score 0 · Answer 2 · answered Mar 25 '23 at 03:58

Alternative approach:

The problem can be solved by Geometry, since the distance function in $~\Bbb{R^2}~$ analogizes to the norm function $~|~x + iy ~|~$ in $~\Bbb{C}.$

In $~\Bbb{R^2},~$ $~\displaystyle \sqrt{x^2 + y^2}~$ represents the distance of the shortest path (i.e. a straight line segment) from $~(x,y)~$ to $~(0,0).~$

Alternatively, $~|x| + |y|~$ represents the taxicab geometry distance of $~(x,y)~$ to $~(0,0).$

Since, in Plane Geometry, the shortest distance between two points is the length of the line segment between the two points, the taxicab geometry distance can never be shorter than the straight line distance.

inequality of modulus of a complex number

2 Answers2