Complex number related problem

Question

Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______

My solution is as follow

${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ & $Z = {z_1} + {z_2} + {z_3} = 2\left( {{e^{i{\theta _1}}} + {e^{i{\theta _2}}} + {e^{i{\theta _3}}}} \right)$

$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {{e^{i{\theta _1}}} - {e^{i{\theta _2}}}} \right| = \left| {{e^{i{\theta _1}}} - {e^{i{\theta _3}}}} \right|$

Let ${\theta _1} = 0$

$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {1 - \left( {\cos {\theta _2} + i\sin {\theta _2}} \right)} \right| = \left| {1 - \left( {\cos {\theta _3} + i\sin {\theta _3}} \right)} \right|$

$ \Rightarrow \left| {1 - \cos {\theta _2} - i\sin {\theta _2}} \right| = \left| {1 - \cos {\theta _3} - i\sin {\theta _3}} \right| \Rightarrow \left| {2{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| {2{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right|$

$\Rightarrow \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right| \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {\cos \frac{{{\theta _2}}}{2} + i\sin \frac{{{\theta _2}}}{2}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {\cos \frac{{{\theta _3}}}{2} + i\sin \frac{{{\theta _3}}}{2}} \right)} \right|$

$ \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right|$

$ \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right|$

$\Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|$

${\theta _2} \ne {\theta _3}$

How do I proceed further?

Answer 1

Consider $$u=z_2/z_1,v=z_3/z_1$$ and then we have $$|u-1|=|v-1|\tag{1}$$ and $$u\neq v, |u|=|v|=1\tag{2}$$ and $$|1+u+v|=1\tag{3}$$ Then from first equation we get $$(u-1)(\bar{u} - 1)=(v-1)(\bar{v}-1)$$ or $$u+\bar{u}=v+\bar{v}\tag{4}$$ This means that $u, v$ have same real part and using $(2)$ we can see that they are conjugates. If $u=x+iy, v=x-iy$ then from $(3)$ we get $$|1+2x|=1$$ so that either $x=0$ or $x=-1$. But $x=-1$ gives $y=0$ so that $u=v$ which is not allowed.

Thus $u, v=\pm i$ and the expression whose value is to be found is $$|z_1|^2|1+u||1+v|=4|1+i||1-i|=8$$

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The key to this problem is to understand that the individual values of $z_1,z_2,z_3$ don't matter, but what matters is their ratios. Also the choice of dividing by $z_1$ to get $u, v$ is because all the constraints in question are symmetric in $z_2,z_3$ and $z_1$ plays a special role as being equidistant from both $z_2,z_3$.

Answer 2

We have three points on the circle of radius $2$ centered at the origin. We know that the triangle formed by these points is isosceles. We can rotate the triangle by some angle so that point $z_1$ is on either the real axis or the imaginary axis. Note that the modulus will not change because rotation = multiplying by $e^{iθ}$ for some $θ$. Now the point $z_2,z_3$ are symmetric about the axis you chose; what can you deduce about their sum?

Spoilers

After the rotation $z_1$ is either $\pm 2$ or $\pm 2i$. WLOG assume it's $2$. Then the rotated $z_2$, $z_3$ are $x+iy$ and $x-iy$. We have $|2+2x|=2$ meaning $x=0$. Thus, $z_2=\pm 2i$ and $z_3=\mp 2i$. Thus, $|z_1+z_2||z_1+z_3|=8$.

Answer 3

This is a kludgy metacheating simplification of the problem. From the constraints, all 3 points are on the circle of radius 2, centered at the origin.

Further, directly from the constraints, $z_1$ is on the perpendicular bisector of the line segment connecting $z_2$ and $z_3$. Finally, $z_2, z_3$ must be chosen so that $|z_1 + z_2 + z_3|$ is also on the same circle. This suggests that the vectors $z_2, z_3$ should cancel each other out, which suggests that $z_2 + z_3$ must equal $0$.

Initially, I considered $(z_2, z_1, z_3) = (-2, 2i, 2)$ as one obvious way of satisfying these constraints. Then, I realized, based on someone else's (now deleted) comment, that this answer could be harmlessly rotated by any angle $\theta$.

Although I can't prove it, it seems probable to me that the only way to satisfy these constraints is by some rotation of either $(z_2, z_1, z_3) = (-2, 2i, 2)$ or $(z_2, z_1, z_3) = (-2, -2i, 2).$

This means that (for example) the triangle formed by the vertices $(0,0), z_1, z_2$ must be a 45-45-90 right triangle, whose hypotenuse is $|z_1 + z_2| = 2\sqrt{2}$.

Consideration between $z_1$ and $z_3$ is identical.

Therefore, the answer is $\left[2\sqrt{2}\right]^2 = 8.$

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Note
A case may be made that this answer is incomplete, since I did not actually prove that there was no other way to satisfy the constraints. I am therefore relying heavily on the fact that the tone of the question indicates that the answer must be unique.

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Addendum
Completing the problem, by proving (for example) that when $z_1 = 2i$ that $z_2, z_3$ must be on the real axis.

I started to analyze this, and then realized that I am merely repeating the analysis already supplied in the answer given by justadzr.

That is, any given $(z_2, z_1, z_3)$ that satisfies the constraints continues to satisfy the constraints when all three points are rotated by $\theta$. When such a rotation takes $z_1$ to $2i$, then, very similar to what justadzr's answer indicates, $z_2, z_3$ must have form $x + iy, -x +iy$, which subsequently implies that $y = 0$.

This places $z_2, z_3$ on the real axis, which places them at $-2, +2$ (in some order), since these are the only two places where the circle intersects the real axis.

Answer 4

Algebra + Geometry approach

Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______

Let's try interpret the problem geometrically.

$|z_1| = |z_2 | = |z_3|$ means that the three points lie on a circle.

$|z_1 - z_2 | = |z_1 - z_3|$ means that the chord substended by $z_1$ and $z_2$ is equal to the chord subtended by $z_1$ and $z_3$. This means that $z_1$ is the mid point of the arc on the circle starting at $z_3 $ and ending at $z_2$.

It is also given that $|z_1 + z_2 + z_3 |= 2$, we can rearrange this to give: $|1 + \frac{z_2}{z_1} +\frac{z_3}{z_1} | = 1$ but we know that $\frac{z_2}{z_1} = e^{i \theta}$ and $\frac{z_3}{z_1} = e^{- i \theta}$, this leads to:

$$| 1+ 2 \cos \theta | = 1$$

It is clear the the only two principle solutions for $\theta$ are $\theta = \{\pi, \frac{\pi}{2} \}$... but if $\theta=\pi$ then $z_2 = z_3$ violating the condition in question (draw a picture!)

We go back to the question : $|z_1 + z_2 | |z_1 + z_3| = \frac{|z_1^2 - z_2^2| | z_1^2 - z_3^2|}{|z_1 - z_2| |z_1 - z_3|}$

We know when we square a complex number, we square the magnitude and double the angle. Hence, $z_1^2 - z_2^2$ represents a chord having twice the angle as $z_1 - z_2$ and on a circle with squared the radius and similarly does $|z_1 - z_3|^2$.

Ok, but what is the expression for the length of chord which subtends an angle $\theta$ in a circle of radius $r$? By the law of cosines, I find it as $ 2r \sin \frac{\theta}{2}$... now we begin the bash:

$$|z_1 + z_2 | |z_1 + z_3| = \frac{|z_1^2 - z_2^2| | z_1^2 - z_3^2|}{|z_1 - z_2| |z_1 - z_3|} = \frac{ (2r^2 \sin \theta)^2}{(2r \sin \frac{\theta}{2})^2}=4r^2 \cos^2 \frac{\theta}{2}$$

We know $r=2$ and $ \theta= \frac{\pi}{2}$, which gives the final answer as $8$

Answer 5

WLOG, it can be assumed that $z_1$ is on Imaginary axis(or y-axis).

Since $\left|\frac{z_1+z_2+z_3}{3}\right|=\frac{2}{3}$

It means that distance of the Centroid of triangle formed by joining $z_1,z_2,z_3$ from the centre of circle is $\frac{2}{3}$ and consequently distance of centroid from $z_1$ is $\frac{4}{3}$.

So perpendicular distance of $z_1$ from line segment joining $z_2$ and $z_3$ is

$\frac{1}{2}(\frac{4}{3})+\frac{4}{3}=2$

(since centroid divides median in ratio 2:1)

which is radius of the circle on which $z_1,z_2,z_3$ lie.

Thus $z_2$ and $z_3$ lie on the x-axis and are thus end-points of the diameter of the circle.

So clearly $z_2=-z_3$ and $z_3=-z_2$.

So, $|z_1+z_2||z_1+z_3|=|z_1-z_3||z_1-z_2|=(2\sqrt{2})(2\sqrt{2})=8$

Answer 6

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 $(3)$