If $|z_1+z_2+z_3|=|z_2+z_3|=|z_1|$ find $\frac{z_1}{z_2+z_3}$

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?

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

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

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

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 $$z_4=e^{\pm\frac{2\pi i}3}=-\frac12\pm i\frac12\sqrt3=\frac{z_1}{z_2+z_3}$$

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 $$(k+1)^2=k\implies...$$

• I don't think this is correct. $z_1=\pm(z_2+z_3)$ does not satisfy all the conditions.
– user750196
Feb 16 '20 at 18:22
• You seem to assume that $|z_1|+|z_2+z_3|=|z_1+z_2+z_3|$, but that is not true.
– user750196
Feb 16 '20 at 18:23

WLOG let $$z_2+z_3=re^{iu}$$ and $$z_1=Re^{iv}$$ where $$r,R\ge0$$ and $$u,v$$ are real

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

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

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

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

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