# Find value of $|z_1+z_2||z_1+z_3|$ Provided $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2\;\;$. and $|z_1-z_3|=|z_1-z_2|\; \;$

Let $$z_1,\;z_2,\;z_3\;$$ be complex number such that $$|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2\;\;$$. If $$|z_1-z_3|=|z_1-z_2|\; \;$$ and $$z_2 \neq z_3.\; \;$$ Then Find value of $$|z_1+z_2||z_1+z_3|$$.

My Thinking:

All I could think is that $$z_1, z_2, z_3 \;$$ lies on a circle of radius $$2$$ with origin as center. Can anyone help me in how to process further.

• Hint: the problem is invariant to rotations $z \mapsto \omega z$ with $|\omega|=1$ and point reflections across the origin $z \mapsto -z$, so it can be assumed WLOG that $z_1 = 2$. Then think of the geometry of it.
– dxiv
May 26, 2022 at 5:43
• @user2661923 You want to look for complex numbers with this approach. I don't think there is a solution to all constraints where $z_1, z_2, z_3$ are all real. If this is a math problem and not just a puzzle you also need to prove uniqueness afterwards. It can still be a useful technique though. May 26, 2022 at 11:38
• @user2661923 Last condition is $z_2 \neq z_3$. I came up with the same examples but this condition (together with the other ones) cannot be satisfies with a real example. May 26, 2022 at 12:43
• @quarague Good point. I have deleted my comments. I never looked at the posting, because it seemed as if the title was showing all of the constraints. My mistake. May 26, 2022 at 12:51
• For what it's worth, it seems to me that the three points must be on the same circle, that $z_1$ must be in between the other two points, and that the angle between $z_1$ and $z_2$ must match the angle between $z_1$ and $z_3$, in order for $|z_1 - z_2|$ to equal $|z_1 - z_3|.$ May 26, 2022 at 13:12

$$|z_1-z_2|=|z_1-z_3|$$ implies $$z_1$$ moves on the perpendicular line bisector of the segment joinng points $$z_2$$ and $$z_3$$. Let $$z_2=2, z_3=-2$$ and $$z_1=2i$$ Hence, $$|z_1+z_2||z_1+z_3|=|2i+2||2i-2|=8.$$

The following formulation seems useful: Let $$z_1,z_2,z_3,z_4$$ be four points on a circle centered at the origin. If $$z_1+z_2+z_3+z_4=0,$$ then up to reindexing, one has $$z_1=-z_3~{\rm and~}z_2=-z_4.$$

This one Is my Own solution

Let $$z_1=2\cdot e^{i \alpha}, z_2=2\cdot e^{i \beta}, z_3=2\cdot e^{i \gamma}$$

and It is given that $$z_2\neq z_3$$ so we must not get $$\beta=\gamma+2k\pi$$

$$z_1+z_2+z_3=2(e^{i\alpha}+e^{i\beta}+e^{i\gamma})\implies |z_1+z_2+z_3|=2|e^{i\alpha}+e^{i\beta}+e^{i\gamma}|$$

Hence $$|e^{i\alpha}+e^{i\beta}+e^{i\gamma}|=1\implies |(\cos\alpha+\cos\beta+\cos\gamma)+i(\sin \alpha+\sin\beta+\sin\gamma)|=1\implies 3+2 \cos(\alpha-\beta)+2\cos(\beta-\gamma)+2\cos(\gamma-\alpha)=1$$

Hence $$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=-1.........(1)$$

From $$|z_1-z_3|=|z_1-z_2|\implies |2e^{i\alpha}-2e^{i\gamma}|=|2e^{i\alpha}-2e^{i\beta}|\implies |(\cos\alpha-\cos\gamma)+i(\sin\alpha-\sin\gamma)|=|(\cos\alpha-\cos\beta)+i(\sin\alpha-\sin\beta)|$$

$$\implies 2-2\cos(\alpha-\gamma)=2-2\cos(\alpha-\beta)$$

$$\implies \cos(\alpha-\gamma)=\cos(\alpha-\beta).......(2)$$

$$\implies \alpha-\gamma=\beta-\alpha \implies2\alpha=\beta+\gamma..........(3)$$

Now using $$(2), (3)$$ Equation $$(1)$$ turns to

$$\cos(\alpha-\beta)+\cos(2\alpha-\gamma-\gamma)+\cos(\alpha-\beta)=-1.......(4)$$

And using the $$\alpha-\gamma=\beta-\alpha$$ again in Equation $$(4)$$ turns to

$$2\cos(\alpha-\beta)+\ \cos(2(\beta-\alpha))=-1\implies \cos(\alpha-\beta)=0$$ or $$-1$$

If $$\cos(\alpha-\beta)=-1$$ then $$\alpha-\beta=(2k+1)\pi\;$$ and from $$(3)$$ we get $$\beta=\gamma+2k\pi\;$$

So $$\cos(\alpha-\beta)$$ must be $$=0$$

Now According to Question

$$|z_1+z_2||z_1+z_3|\implies |z_1^2+z_1z_2+z_1 z_3+z_2z_3|$$

$$\implies \;4|e^{i2\alpha}+e^{i(\alpha+\beta)}+e^{i(\alpha+\gamma)}+e^{i(\beta+\gamma)}|$$

$$\implies$$ $$4|e^{2i\alpha}+\cos(\alpha+\beta)+i\sin(\alpha+\beta)+\cos(\alpha+\gamma)+i\sin(\alpha+\gamma)+e^{2i\alpha}|$$

$$\implies 4\bigg|2\cdot e^{2i\alpha}+2\cos\bigg(\dfrac{2\alpha+\beta+\gamma}{2}\bigg)\cos\bigg(\dfrac{\beta-\gamma}{2}\bigg)+i\cdot 2\sin\bigg(\dfrac{2\alpha+\beta+\gamma}{2}\bigg)\cos\bigg(\dfrac{\beta-\gamma}{2}\bigg)\bigg|$$

$$\implies$$ $$4\bigg|2\cdot (\cos2\alpha+i\sin2\beta)+2\cos\bigg(\dfrac{\beta-\gamma}{2}\bigg)(\cos2\alpha+i\sin2\beta)\bigg|$$

$$\implies 8\bigg|\bigg(\cos2\alpha+i\sin2\beta\bigg)\bigg(1+\cos\bigg(\dfrac{\beta-\gamma}{2}\bigg)\bigg|$$

$$\implies 8\bigg| \bigg(1+\cos\bigg(\dfrac{\beta-\gamma}{2}\bigg) \bigg |$$

$$\implies8\bigg| \bigg(1+\cos(\alpha-\beta)\bigg) \bigg |=8$$

Note:

$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$

$$z_{1}^2=4\cdot e^{2i\alpha},\;z_{1}z_{3}=4\cdot e^{i(\alpha+\gamma)},\;z_{1}z_{2}=4\cdot e^{i(\alpha+\beta)},\;z_{3}z_{2}=4\cdot e^{i(\gamma+\beta)}$$

$$\cos \alpha +\cos \beta=2\cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\cos\bigg(\dfrac{\alpha-\beta}{2}\bigg)$$

$$\sin \alpha +\sin \beta=2\sin\bigg(\dfrac{\alpha+\beta}{2}\bigg)\cos\bigg(\dfrac{\alpha-\beta}{2}\bigg)$$

$$\beta-\gamma=2(\alpha-\beta)$$ using Equation $$(3)$$