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In my complex analysis class, we went over a geometric proof of this with the triangle inequality, but I'm trying to find a more algebraic proof. I'm also trying not to use Arg because we haven't gone over it in my class yet and all I really know about an argument is that if $z_1$, $z_2$, and 0 are collinear than they'd all have the same angle off the real axis.

I haven't really gotten particularly far. I can convert the absolute values (based on the definition of modulus that, for a complex number $z=x+iy$, $|z|=sqrt{(x^2+y^2)}$). I can then square both sides of the equation twice and then simplify to get something that looks really easy to work with, but I'm not sure what the results of that could tell me about how the three points are collinear.

Any help would be much appreciated.

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  • $\begingroup$ Actually antiparallel vectors won't satisfy the equation (unless you consider 0 and any complex number to be antiparallel). $\endgroup$ Jan 29 '19 at 22:03
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Say $z_1 = a+ib$ and $z_2 = c+id$, which correspond to the vectors $(a,b)$ and $(c,d)$ in $\mathbb{R}^2$. Their norms squared are $\lvert z_1 \rvert^2 = a^2+b^2$ and $\lvert z_2 \rvert^2=c^2+d^2$. Now $$z_1+z_2 = a+c+i(b+d),$$ hence its norm squared is $$\lvert z_1+z_2\rvert^2 = (a+c)^2+(b+d)^2 = \lvert z_1\rvert^2 + \lvert z_2 \rvert^2+2(ac+bd).$$ Moreover $$ \lvert z_1z_2\rvert = \sqrt{(a^2+b^2)(c^2+d^2)}. $$ Now, your equation is equivalent to $$\lvert z_1+z_2\rvert^2 = \left(\lvert z_1\rvert+\lvert z_2\rvert\right)^2=\lvert z_1\rvert^2+\lvert z_2\rvert^2+2\lvert z_1z_2\rvert$$ which implies $$ac+bd=\sqrt{(a^2+b^2)(c^2+d^2)},$$ namely $(ad-bc)^2=0$. So $ad-bc=0$, i.e. $z_1$ is parallel to $z_2$. You can see this as follows: if $\langle \cdot, \cdot \rangle$ denotes the standard scalar product in $\mathbb{R}^2$, then $ad-bc=0$ is the same as $\langle (a,b), (d,-c)\rangle=0$. But $\langle (d,-c),(c,d)\rangle = 0$, which means that $z_2$ is orthogonal to $(d,-c)$, which is in turn orthogonal to $z_1$. So finally $z_1$ and $z_2$ are parallel.

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  • $\begingroup$ Thank you so much! $\endgroup$
    – Masie
    Jan 30 '19 at 2:47
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If $z_1$ is a positive real number, what does this look like? $$|z_1+(x_2+y_2i)|=|z_1|+|x_2+y_2i|=z_1+\sqrt{x_2^2+y_2^2}$$ $$|z_1+(x_2+y_2i)| = \sqrt{(z_1+x_2)^2+y_2^2}$$ Equate the two, and square them: $$z_1^2+x_2^2+y_2^2+2z_1\sqrt{x_2^2+y_2^2} = \left(z_1+\sqrt{x_2^2+y_2^2}\right)^2 = (z_1+x_2)^2+y_2^2 = z_1^2+x_2^2+y_2^2 + 2z_1x_2$$ $$2z_1\sqrt{x_2^2+y_2^2}=2z_1x_2$$ So then, $x_2=\sqrt{x_2^2+y_2^2}\ge |x_2|$. Equality there is only possible if $y_2=0$, and equality of $x_2$ and $|x_2|$ is then only possible if $x_2\ge 0$. So that's it - for positive real $z_1$, $|z_1+z_2|=|z_1|+|z_2|$ only if $z_2$ is a nonnegative real number. Either one of them's zero, or they're parallel.

Of course, we would like something that works for general $z$. For that, consider what happens if we multiply both $z_1$ and $z_2$ by the same unit vector $w$: $|z_1w|=|z_1|\cdot |w|=z_1$ and $|z_1w+z_2w|=|(z_1+z_2)w|=|z_1+z_2|\cdot |w|=|z_1+z_2|$. None of the absolute values change, so we can multiply a solution by $w$ to get another solution. In particular, if $|z_1+z_2|=|z_1|+|z_2|$ (and they're nonzero), then by taking $w=\frac{|z_1|}{z_1}$, we get a new solution $z_1',z_2'$ with $z_1'=|z_1|$ a positive real number. By our previous work, this makes $z_2'$ also positive real. Undo the rotation: $z_1=\frac1w\cdot z_1'$ and $z_2=\frac1w\cdot z_2'$ are both positive real multiples of the same unit vector $\frac1w$, so they're parallel. Victory - equality in the triangle inequality means they're parallel.

OK, there's also the case when one or both of the vectors is zero. Let's just say that zero is parallel to everything, and be done with that.

What about the "antiparallel" in your statement? That has to do with equalities like $|z_1+z_2|=|z_1|-|z_2|$ or $|z_1+z_2|=|z_2|-|z_1|$.

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Let’s assume

$$z_1=x_1+iy_1$$ $$z_2=x_2+iy_2$$

Computing one gets

$$|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=x_1^2+y_1^2+x_2^2+y_2^2+2(x_1x_2+y_1y_2)$$

And

$$(|z_1|+|z_2|)^2=x_1^2+y_1^2+x_2^2+y_2^2+2|z_1z_2|$$

The identity we are investigating is therefore equivalent to

$$|z_1z_2|=(x_1x_2+y_1y_2)$$

Equivalently

$$|z_1z_2|^2=(x_1x_2+y_1y_2)^2$$

Now $z_1z_2=x_1x_2-y_1y_2+i(x_1y_2+x_2y_1)$ which means that

$$|z_1z_2|^2=(x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2$$

After expanding and identifying, one gets

$$(x_1y_2)^2+(x_2y_1)^2-2x_1x_2y_1y_2=(x_1y_2-x_2y_1)^2=0$$

And this tells us that $x_1y_2-x_2y_1=0$ which is equivalent to $z_1$ and $z_2$ are colinear

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I will answer the question in the heading (not including $0$). Let $z_k=x_k+iy_k$ Square both sides and remove common terms to get: $(x_1+iy_1)(x_2-iy_2)+(x_1-iy_1)(x_2+iy_2)=(2(x_1x_2+y_1y_2))=2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$. Now square both sides and eliminate common terms to get $2x_1x_2y_1y_2=x_1^2y_2^2+x_2^2y_1^2$ Assume none of the terms $=0$, divide by the product and get $2=\frac{x_1y_2}{x_2y_1}+\frac{x_2y_1}{x_1y_2}$ Notice that the two terms on the right are reciprocal, su that with $u=$ one ratio, we have a quadratic $u^2-2u+1=0$, so that the ratio $=1$, forcing the vectors to be scalar multiples of each other.

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Let $z_1=r_1e^{i\theta_1}=r_1\cos\theta_1+ir_1\sin\theta_1, z_2=r_2e^{i\theta_2}=r_2\cos\theta_2+ir_2\sin\theta_2$.

Then $\mid z_1+z_2\mid=\mid (r_1\cos\theta_1 +r_2\cos\theta_2)+i(r_1\sin\theta_1 +r_2\sin\theta_2)\mid=\sqrt{r_1^2\cos^2\theta_1 +2r_1r_2\cos\theta_1\cos\theta_2+r_2^2\cos^2\theta_2+r_1^2\sin^2\theta_1 +2r_1r_2\sin\theta_1 \sin\theta_2 +r_2^2\sin^2\theta_2}=\sqrt{r_1^2+r_2^2+2r_1r_2(\cos\theta_1, \sin\theta_1) \cdot (\cos\theta_2,\sin\theta_2) }=r_1+r_2\implies (\cos\theta_1, \sin\theta_1) \cdot (\cos\theta_2, \sin\theta_2)=1\implies \theta_1 =\pm\theta_2 $.

Now I know you said you didn't want to use "arg", but all I used was polar coordinates. And the statement is really about the arguments of the two numbers.

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  • $\begingroup$ I guess I should have specified that I'm not quite comfortable using Euler's formula and exponential form as well, but this is quite clever and I do understand. $\endgroup$
    – Masie
    Jan 30 '19 at 2:48
  • $\begingroup$ Great. I figured I probably shouldn't have done it. Rather, I should have tried with just $z=x+iy$. Maybe this can be done. But you will learn in short order that a complex number can be specified by an angle and a modulus, as well as the aforementioned Euler's formula, both of which are extremely useful. Glad you understand. $\endgroup$
    – user403337
    Jan 30 '19 at 3:07

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