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

Question

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

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

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 $a\overline{b} = (\alpha \gamma + \beta\delta) + i(\beta \gamma - \alpha\delta)$, which tells us that $$ \left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right) \iff (\alpha \gamma + \beta \delta \ge 0) \ \ \wedge \ \ (\alpha\delta= \beta \gamma ) $$ From here, I do the following \begin{align} &2(\alpha\delta)^2 = 2(\alpha\delta)^2 \iff 2(\alpha\delta)(\alpha\delta) = (\alpha\delta)^2 + (\alpha\delta)^2 \iff 2\alpha\delta\beta \gamma = (\alpha\delta)^2 + (\beta \gamma)^2 \notag \\ \iff& (\alpha\gamma)^2 + 2\alpha\gamma\beta \delta + (\beta \delta)^2 =(\alpha\gamma)^2 + (\alpha\delta)^2 + (\beta \gamma)^2 + (\beta \delta)^2 \iff (\alpha\gamma + \beta \delta)^2 = \left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right) \notag \\ \iff& 2(\alpha\gamma + \beta \delta) = 2\sqrt{\left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right)} \qquad \ \text{(here using the hypothesis that $\alpha \gamma + \beta \delta \ge 0$)} \notag \\ \iff & \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha\gamma + \beta \delta) = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 +2\sqrt{\left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right)}\notag\\ \iff& \left(\alpha^2 +2\alpha\gamma + \gamma^2 \right)+ \left(\beta^2 +2 \beta \delta+ \delta^2\right) = \left(\sqrt{\alpha^2 + \beta^2}\right)^2 +2\sqrt{\alpha^2 + \beta^2}\sqrt{\gamma^2 + \delta^2} + \left(\sqrt{\gamma^2 + \delta^2}\right)^2\notag\\ \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!

Answer 1

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 $|z|^2 = z\overline z$. So maybe it’s easier to prove equivalence to the squared identity. As others already have hinted, $|a + b|^2 = |a|^2 + 2\operatorname{Re} a\overline b + |b|^2$. Hence

\begin{align*} |a + b|^2 = (|a| + |b|)^2 &\iff 2\operatorname{Re} a\overline b = 2|a||b| \\ &\iff \operatorname{Re} a\overline b = |a\overline 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.

Answer 2

This could be useful:

$$ |a + b|^2 = |a|^2 +|b|^2 + 2 \mathrm{Re} (a\bar{b}) $$

Answer 3

HINT:

Note that

$$|a+b|^2=|a|^2+|b|^2+2\text{Re}(a\bar b)$$

while $(|a|+|b|)^2=|a|^2+|b|^2+2|a||b|$

Answer 4

Here's a geometric approach. Consider $a$ and $b$ as vectors in the complex plane and use the representation $a = \alpha e^{i\theta}$, $b = \beta e^{i\phi}$, 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, $a\overline{b} = \alpha\beta e^{i(\theta-\phi)}$, which is positive iff $\theta = \phi$.