Prove that $|a + b| = |a| + |b| iff aoverline{b} ge 0$

Question

I'm reading a complex analysis book. In this book, the author establishes the following statement

If $a,b in mathbb{C}$, then $|a + b| = |a| + |b|  iff  left(aoverline{b}in mathbb{R}right) wedge left(aoverline{b}ge 0right)$

This statement didn't seem intuitive to me, so I decided to try to prove it. I denoted the complex numbers $a$ and $b$ as $a = alpha + i beta$ and $b = gamma + i delta$. Using this, I get that $aoverline{b} = (alpha gamma + betadelta) + i(beta gamma - alphadelta)$, which tells us that
$$
 left(aoverline{b}in mathbb{R}right) wedge left(aoverline{b}ge 0right) iff (alpha gamma + beta delta ge 0)    wedge   (alphadelta= beta gamma )
$$
From here, I do the following
begin{align}
&2(alphadelta)^2 = 2(alphadelta)^2 iff 2(alphadelta)(alphadelta) = (alphadelta)^2 + (alphadelta)^2 iff 2alphadeltabeta gamma = (alphadelta)^2 + (beta gamma)^2 notag \
iff& (alphagamma)^2 + 2alphagammabeta delta + (beta delta)^2 =(alphagamma)^2 +  (alphadelta)^2 + (beta gamma)^2 + (beta delta)^2 iff (alphagamma + beta delta)^2 = left(alpha^2 + beta^2right)left(gamma^2 + delta^2right) notag \
iff& 2(alphagamma + beta delta) = 2sqrt{left(alpha^2 + beta^2right)left(gamma^2 + delta^2right)} qquad  text{(here using the hypothesis that $alpha gamma + beta delta ge 0$)} notag \
iff & alpha^2 + beta^2 + gamma^2 + delta^2 + 2(alphagamma + beta delta) = alpha^2 + beta^2 + gamma^2 +  delta^2 +2sqrt{left(alpha^2 + beta^2right)left(gamma^2 + delta^2right)}notag\
iff& left(alpha^2 +2alphagamma + gamma^2 right)+ left(beta^2 +2 beta delta+ delta^2right)  = left(sqrt{alpha^2 + beta^2}right)^2 +2sqrt{alpha^2 + beta^2}sqrt{gamma^2 + delta^2} + left(sqrt{gamma^2 +  delta^2}right)^2notag\
iff& (alpha + gamma)^2 + (beta + delta)^2 = left(sqrt{alpha^2 + beta^2} +sqrt{gamma^2 +  delta^2}right)^2 iff |a+ b|^2 = left(|a| + |b|right)^2 iff  |a+ b| = |a| + |b|
end{align}
where in the last equivalence I used the fact that $|z|ge 0,  forall z in mathbb{C}$.
Is my proof correct? And also, does anyone know a different (possibly shorter) method of proving the above statement? Any and all help would be greatly appreciated. Thank you!

k.stm · Answer

First off: Intuitively the statement reads “The distances to the origin of two complex points $a$ and $b$ add up to the distance to the origin of $a + b$ if and only if they lie on the same ray from origin.” This is because $overline b$ is $b$ reflected on real line, which can be interpreted as “$b$, just with its angle to the positive real ray inverted”.
Next, the absolute value and complex conjugation are related by $|z|^2 = zoverline z$. So maybe it’s easier to prove equivalence to the squared identity. As others already have hinted, $|a + b|^2 = |a|^2 + 2operatorname{Re} aoverline b + |b|^2$. Hence
begin{align*}
|a + b|^2 = (|a| + |b|)^2 &iff 2operatorname{Re} aoverline b = 2|a||b| \
&iff operatorname{Re} aoverline b = |aoverline b|.
end{align*}
So this reduces to proving for $z ∈ ℂ$, $operatorname{Re} z = |z| iff z ∈ [0..∞)$, which shouldn’t be hard to do.

Phil · Answer

Here's a geometric approach. Consider $a$ and $b$ as vectors in the complex plane and use the representation $a = alpha e^{itheta}$, $b = beta e^{iphi}$, for $alpha,beta$ real and positive. Then it should be geometrically clear that $|a|+|b| = |a+b|$ iff $a$ and $b$ have the same direction, i.e., $theta = phi$; this is also easy to show algebraically.
On the other hand, $aoverline{b} = alphabeta e^{i(theta-phi)}$, which is positive iff $theta = phi$.

Mark Viola · Answer

HINT:
Note that
$$|a+b|^2=|a|^2+|b|^2+2text{Re}(abar b)$$
while $(|a|+|b|)^2=|a|^2+|b|^2+2|a||b|$

ir7 · Answer

This could be useful:
$$ |a + b|^2 = |a|^2 +|b|^2 + 2 mathrm{Re} (abar{b}) $$

Prove that $|a + b| = |a| + |b| iff aoverline{b} ge 0$

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