I'm asked to prove that if $z_1$, $\ldots$, $z_k$ lie on one side of a straight line through $0$, then $z_1+\cdots+z_k \neq 0$.

In the proof, we let $\theta$ be the angle between the line and the real axis, and let $w = \cos\theta + i\sin\theta$. Then $z_1w^{-1}$, $\ldots$, $z_kw^{-1}$ all lie on one side of the real axis, so the same is true for $z_1w^{-1}+\cdots+z_kw^{-1} = (z_1+\cdots+z_k)w^{-1}$, which shows that $z_1+\cdots+z_k$ lies on the corresponding side of the original line.

But why is it the case that if we divide the various $z_i$s of the original sequence by $w$ then they all get sorted on one side of the real axis? I get that the points all rotate clockwise by $\theta$, but why?

Thank you all in advance.


Edit: I think I've answered it myself; see below.

  • $\begingroup$ Multiplying by $w^{-1}$ rotates everything clockwise by $\theta$, so it map the line through $0$ and $w$ to the real axis. $\endgroup$ – kccu Apr 17 at 14:41
  • $\begingroup$ "Multiplying by $w^{-1}$ rotates everything clockwise by θ" Sure, but why? $\endgroup$ – Simone Apr 17 at 14:45
  • 1
    $\begingroup$ @Simone Because multiplying be $e^{i\theta}$ in the complex plane is equivalent to a (counter-clockwise) rotation of angle $\theta$ and center $O$. Why, you may still ask? Well, just write the complex product and compare to the rotation in $\Bbb R^2$. See also this. $\endgroup$ – Jean-Claude Arbaut Apr 17 at 16:07
  • $\begingroup$ "I get that the points all rotate clockwise by theta, but why?" I'm having trouble parsing this. Do you mean that you accept that the statement is true, but don't know why it's true? (The reason I ask is that usually, "get" in this context means "understand," not simply "accept.") $\endgroup$ – Brian Tung Apr 17 at 16:19
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    $\begingroup$ Note that there's nothing special about the complex plane here - it is just a dimension $2$ vector space. In general it's true that if $v_1, \ldots, v_k \in \mathbb{R}^n$ are all to one side of a hyperplane $H$ then the sum will also be on that side (and in particular, will not be $0$.) $\endgroup$ – Jair Taylor Apr 17 at 16:44


It is not difficult to prove the case $k=2$, than use induction.

Or, following your way:

if all $z_i$ lie on the same side of the line $z=\rho e^{i\theta}$ we have
$$z_i=\rho_ie^{i(\theta +\alpha_i)}$$ with $0<\alpha_i<\pi$ or $-\pi<\alpha_i<0$.

So we have:

$z_i e^{-i\theta}=\rho_ie^{i(\theta +\alpha_i)}e^{-i\theta}=\rho_ie^{i \alpha_i}$

which, due to the limits for $\alpha_i$, are all on the same side of the real axis.

  • $\begingroup$ I don't feel like this answers the question; I have already outlined the proof. $\endgroup$ – Simone Apr 17 at 14:58

I might have convinced myself; but should I be convinced?

I suspect that it's easier for me to think of this points using polar coordinates: we started with: $$x+iy=z=|z|(\cos\theta+i\sin\theta)$$ so $$x=|z|\cos\theta$$and$$y=|z|\sin\theta$$ in a comment below my initial question I've found that the $\theta$ clockwise rotation of $z$ obtained by $zw^{-1}$ as above is given in terms of $x$ and $y$ by the formula: $$z''=(x\cos\theta+y\sin\theta)-i(x\sin\theta-y\cos\theta)$$ let $x\gt0$ and we can accept $\theta=arctan (y/x)$, and thus we can substitute as such:

$${x\over \sqrt{y^2+x^2\over x^2}}+{y^2\over x\sqrt{y^2+x^2\over x^2}}-{iyx\over x\sqrt{y^2+x^2\over x^2}}+{iy\over \sqrt{y^2+x^2\over x^2}}={x^2\over\sqrt{y^2+x^2}}+{y^2 \over\sqrt{y^2+x^2}}={x^2\over |z|}+{y^2\over |z|}= \frac 1{|z|}(x^2+y^2)=\frac 1{|z|}{|z|}^2\cos^2\theta+\frac 1{|z|}{|z|}^2\sin^2\theta=|z|(\cos^2\theta+\sin^2\theta)=|z|=|z|(\cos0+i\sin0)$$ which is the point $z''$ with the same modulus $|z|$ of $z$ but laid on the real axis.

makes sense?

  • $\begingroup$ Obviously you can substitute for $x$ and $y$ in $z''$ right away without the substitution of $\theta$, silly me. $\endgroup$ – Simone Apr 17 at 19:57

Got it!

let $$z=|z|(\cos(\theta+|r|)+i\sin(\theta+|r|))$$

so $$zw^{-1}=\frac{|z|(\cos(\theta+|r|)+i\sin(\theta+|r|))}{\cos\theta+i\sin\theta}=|z|\frac{\cos\theta\cos|r|-\sin\theta\sin|r|+i\sin\theta\cos|r|+i\cos\theta\sin|r|}{\cos\theta+i\sin\theta}=|z|\frac{\cos\theta\cos|r|+i^2\sin\theta\sin|r|+i\sin\theta\cos|r|+i\cos\theta\sin|r|}{\cos\theta+i\sin\theta}=|z|\frac{(\cos\theta+i\sin\theta)(\cos|r|+i\sin|r|)}{\cos\theta+i\sin\theta}=|z|(\cos|r|+i\sin|r|)$$

which is the point $z''$ of lenght $|z|$ with an angle of $|r|$ radians.

I feel like this answers the question.


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