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 $|z_1-z_3|=|z_1-z_2|\; \;$

Question

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.

Answer 1

$|z_1-z_2|=|z_1-z_3|$ 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, $|z_1+z_2||z_1+z_3|=|2i+2||2i-2|=8.$

Answer 2

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 $$z_1=-z_3~{\rm and~}z_2=-z_4.$$

Answer 3

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 $(3)$