Proof: $|\alpha + \beta|=|\alpha|+| \beta|$ if and only if $\alpha=0$ or $\beta=0$ or $\alpha/ \beta$ is a real number bigger than $0$.

Question

Proof with complex numbers: $|\alpha + \beta|=|\alpha|+| \beta|$ if and only if $\alpha=0$ or $\beta=0$ or $\alpha/ \beta$ is a real number bigger than $0$.

I don't succeed to proof this statement.

What is the best way to start the proof?

I took: $\alpha=a+ib$ and $\beta=c+id$

Answer 1

Write $a= \alpha$ and $b= \beta$, then after squaring we get:

$$(a+b)(a'+b') = aa'+2|ab|+bb' $$ so $$ab'+a'b = 2|ab|$$

write $x=ab'$ then we have $x+x' = 2|x|$, so $Re(x)=|x|$ so chathethus is equal to hyphottenus if you draw this in complex plane. But this is possible only iff $x\in \mathbb{R}$, so $a/b = x/|b|^2\in \mathbb{R}$

Answer 2

Ahh, this is even easier. We know that for all $a,b$ we have $|a+b|\leq |a|+|b|$ just draw a parallelogram on points $0,a,b$ and $a+b$ in complex plane.

Then if you look at triangle $0,a,a+b$ you see this (triangle) inequality. We will have equality case iff $a$ and $b$ are on line form origin through $a+b$. So $a=kb$ and we rae done.

Answer 3

Let $\alpha=ae^{iA},\beta=be^{iB}$ where $a,b>0$

$$|\alpha+\beta|^2=(a\cos A+b\cos B)^2+(a\sin A+b\sin B)^2=a^2+b^2+2ab\cos(A-B)$$

$$\le a^2+b^2+2ab$$

The equality occurs $ab=0$ or $\cos(A-B)=1\iff A\equiv B\pmod{2\pi}$