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If $z_1,z_2,z_3 \in \mathbb{C}^*$ such that $z_2+z_3\neq 0$ and $|z_1+z_2+z_3|=|z_2+z_3|=|z_1|$, find the value of

$$\frac{z_1}{z_2+z_3}$$

Because $|\frac{z_1}{z_2+z_3}|=1$, the value has to be a complex number on the unit circle. So I did it the dull way with $\frac{z_1}{z_2+z_3}=a+bi$ and I got $a^2+b^2=1$ and $(a+1)^2+b^2=1$ so $2a+1=0$ and in the end there are two possible values:

$$\frac{z_1}{z_2+z_3}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$$

Is there a better way to solve this problem?

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Let $w= \frac{z_2+z_3}{z_1}$. Then, the given condition $|z_1+z_2+z_3|=|z_2+z_3|=|z_1|$ becomes

$$|1+w|=|w|=1$$

Note $|1+w|^2 = 1 + w + \bar w +|w|^2= 2 + w + \frac 1w = 1$, or,

$$w+\frac1w+1=0$$

which is quadratic in $\frac1w$. Solve to obtain

$$ \frac{z_1}{z_2+z_3}=\frac 1w = -\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$$

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  • $\begingroup$ Can you please explain $|1+w|=|w|=1\implies w+\frac1w+1=0$? I don't see it. $\endgroup$
    – user750196
    Feb 16 '20 at 18:35
  • $\begingroup$ @ryan - added one more step in the answer $\endgroup$
    – Quanto
    Feb 16 '20 at 18:38
  • $\begingroup$ Thank you, guys. It's clear now. $\endgroup$
    – user750196
    Feb 16 '20 at 18:39
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Let $z_4=\frac{z_1}{z_2+z_3}$. Then $$|1+z_4|=1=|z_4|$$

If you think about it geometrically, $1$, $z_4$ and $-(1+z_4)$ form the sides of an equilateral triangle in the complex plane. Thus $$z_4=e^{\pm\frac{2\pi i}3}=-\frac12\pm i\frac12\sqrt3=\frac{z_1}{z_2+z_3}$$

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Put $a=z_2+z_3$ and $b=z_1$. Now we have $ |a+b| = |a|=|b|$ and we are looking for $b/a=:k$. Write $$|a||1+k| = |a|\implies |1+k|=1\implies 1+k+k'+kk' =1$$

so $\boxed{(k+1)k'=-k}$. Similary we have $$|k+1| = |k|\implies \boxed{k'=-k-1}$$

Solving this syatem we get $$ (k+1)^2=k\implies...$$

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  • $\begingroup$ I don't think this is correct. $z_1=\pm(z_2+z_3)$ does not satisfy all the conditions. $\endgroup$
    – user750196
    Feb 16 '20 at 18:22
  • $\begingroup$ You seem to assume that $|z_1|+|z_2+z_3|=|z_1+z_2+z_3|$, but that is not true. $\endgroup$
    – user750196
    Feb 16 '20 at 18:23
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WLOG let $z_2+z_3=re^{iu}$ and $z_1=Re^{iv}$ where $r,R\ge0$ and $u,v$ are real

$|z_1|=|z_2+z_3|\implies R=r\implies \dfrac{z_1}{z_2+z_3}=?$

$\implies r=r\sqrt{(\cos u+\cos v)^2+(\sin u\sin v)^2}$

$\iff1=\sqrt{2+2\cos(u-v)}$

$\implies2+2\cos(u-v)=1\iff\cos(u-v)=-\dfrac12=\cos120^\circ$

$u-v\equiv?\pmod{360^\circ}$

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