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Let $z_1,z_2$ be two non-zero complex numbers satisfying $$\lvert z_1+z_2 \rvert=\lvert z_1-z_2 \rvert$$


I have done many things while trying to find this one out.

$\lvert z_1+z_2 \rvert^2= \lvert z_1-z_2 \rvert^2$

$\implies \lvert z_1 \rvert^2 +\lvert z_2 \rvert^2 + 2\lvert z_1 \rvert\cdot \lvert z_2 \rvert\cdot cos(\theta_1-\theta_2)=\lvert z_1 \rvert^2 +\lvert z_2 \rvert^2 - 2\lvert z_1 \rvert\cdot \lvert z_2 \rvert\cdot cos(\theta_1-\theta_2)$

$\implies cos(\theta_1-\theta_2)=0$

$\implies\theta_1-\theta_2 =\dfrac{\pi}{2} $

The three sides of the triangle are :

$ \lvert z_1 \rvert $,$ \lvert z_2 \rvert $,$ \lvert z_1-z_2 \rvert$ respectively.

And I did several other things like I assumed $z_1=x_1+iy_1$ & $z_2=x_2+iy_2$

And by using the relation $\lvert z_1+z_2 \rvert=\lvert z_1-z_2 \rvert$, we get

$x_1x_2=y_1y_2$

I then considered the vertices of the triangle be like $(0,0),(1,2),(2,1)$ where $x_1=1,x_2=2 \text {&} y_1=2,y_2=1$ But this does not give me the answer $\dfrac{ (z_1+z_2)}{2}$

Please give me a generic way to solve this.

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  • $\begingroup$ You proved that the triangle is a right triangle. Geometrically, the circumcenter of a right triangle is the midpoint of its hypotenuse. To complete the proof, take $z = \frac{1}{2}(z_1+z_2)$ and prove that $|z|=\frac{1}{2}|z_1-z_2|\,$. $\endgroup$
    – dxiv
    Commented Apr 20, 2018 at 18:13
  • $\begingroup$ @dxiv Ah! yes yes that is why that answer. Yes. Thanks . No need to answer this. I can do it myself. Should I remove this post? It is needless. So easy! $\endgroup$
    – Saradamani
    Commented Apr 20, 2018 at 18:16
  • $\begingroup$ The geometric hint is fine, but you can also prove it strictly algebraically. With $z=\frac{1}{2}(z_1+z_2)$ and using the given condition, you get that $|z|=|z-z_1|=|z-z_2|$, so $z$ is equidistant from the three vertices, and therefore it is the circumcenter. $\endgroup$
    – dxiv
    Commented Apr 20, 2018 at 18:26

3 Answers 3

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Since $$|z_1- z_2| = |z_1+z_2|$$ we see that $z_1$ is on perpendicular bisector of segment between $z_1$ and $z_2$, so angle $\angle z_10z_2 = 90^{\circ}$ which means that circumcenter is a midpoint of segment between $z_1$ and $z_2$. So $$\zeta ={z_1+z_2\over 2}$$ where $\zeta $ is circumcenter.

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  • $\begingroup$ I have done the same thing . See my answer below. @ChristianF $\endgroup$
    – Saradamani
    Commented Apr 20, 2018 at 19:24
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As I showed this earlier and @dxiv has enlightened my insight,

$\lvert z_1+z_2 \rvert^2= \lvert z_1-z_2 \rvert^2$

$\implies \lvert z_1 \rvert^2 +\lvert z_2 \rvert^2 + 2\lvert z_1 \rvert\cdot \lvert z_2 \rvert\cdot cos(\theta_1-\theta_2)=\lvert z_1 \rvert^2 +\lvert z_2 \rvert^2 - 2\lvert z_1 \rvert\cdot \lvert z_2 \rvert\cdot cos(\theta_1-\theta_2)$

$\implies cos(\theta_1-\theta_2)=0$

$\implies\theta_1-\theta_2 =\dfrac{\pi}{2} $

The triangle is right angled at the origin as $ z_1 $,$z_2$ sides make an angle of $\dfrac{\pi}{2}$ with each other at the origin.

Thus the hypotenuse is $ z_1 - z_2$ and in a right angled triangle, the circumcenter is at the midpoint of the hypotenuse

Thus the circumcentre is $\dfrac{1}{2}.(z_1+z_2)$

And as @dxiv has pointed To complete the proof, take $ z=\dfrac{1}{2}.(z_1+z_2)$ and prove that $\lvert z \rvert= \dfrac{1}{2}.\lvert z_1-z_2 \rvert$

Proved.

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  • $\begingroup$ Nice +1............ $\endgroup$
    – nonuser
    Commented Apr 20, 2018 at 19:26
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Let $z_1=(x_1,y_1)$ and $z_2=(x_2,y_2)$ ans $0 = (0,0,0)$

The circumcenter $C=(c_1,c_2)$ satisfies $$ |C-0|=|C-z_1|= |C-z_2|$$

That us $$ c_1^2+c_2^2 =(c_1-x_1)^2 + (c_2-y_1)^2 =(c_1-x_2)^2 + (c_2-y_2)^2$$

Upon some cancellations we have a system to solve for $c_1$ and $c_2$

$$2c_1x_1+2c_2y_1=x_1^2+y_1^2\\2c_1x_2+2c_2y_2=x_2^2+y_2^2$$

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