Fdtxjr

Given that complex numbers $z_1,z_2,z_3$ lie on unit circle and

$$|z_1-z_2|^2+|z_1-z_3|^2=4$$ Then find value of $|z_2+z_3|$

My try:

We can take $z_1=e^ialpha$, $z_2=e^ibeta$ and $z_3=e^igamma$

So we have

$$|z_1-z_2|=2sinleft(fracalpha-beta2right)$$

$$|z_1-z_3|=2sinleft(fracalpha-gamma2right)$$

So we get:

$$sin^2left(fracalpha-beta2right)+sin^2left(fracalpha-gamma2right)=1$$

Now we have:

$$|z_2+z_3|=2cosleft(fracbeta-gamma2right)$$

Any help here?

As pointed out in comment by Stancek, whom I thank, my previous answer was not correct since it only reported one solution, that was indenpendent on the choice of $z_1$, $z_2$ and $z_3$, i.e
$$|z_2+z_3|=0,$$
easily found by Pythagorean Theorem.

The general solution can be determined as follows. Use the fact that $|z|^2 = zz^*$, to re-write the equation in the form
$$|z_1|^2 - 2Rez_1z_2^* + |z_2|^2 + |z_1|^2 - 2Rez_1z_3^* + |z_3|^2=4.$$
Since the points lie on the unit circle, the equation then simplifies to
$$ Rez_1z_2^* = -Rez_1z_3^*.$$
If you set $z_1 = 1$ you get
$$Rez_2 = -Rez_3.$$
Which yields the two solutions
$$z_3 = -z_2$$
and
$$z_3 = -z_2^*.$$
The first solution leads to
$$boxedz_2+z_3,$$
that is to $z_1$, $z_2$, and $z_3$ being vertices of a right-angled triangle.
The second one, in the case $z_1=1$ can be expressed as
$$|z_2+z_3|= 2left|Imz_2right|.$$
Taking into account the arbitray phase of $z_1$ yields
$$boxed.$$

Note that if $|z|=1$, then $overlinez=1/z$. Thus for $z_1$, $z_2$, and $z_3$ on the unit circle, we have

$$beginalign
|z_1-z_2|^2+|z_1-z_3|^2
&=left(z_1-z_2right)left(1over z_1-1over z_2right)+left(z_1-z_3right)left(1over z_1-1over z_3right)\
&=4-left(z_1over z_2+z_2over z_1+z_1over z_3+z_3over z_1 right)\
&=4-left((z_1^2+z_2z_3)(z_2+z_3)over z_1z_2z_3 right)
endalign$$

It follows that $|z_1-z_2|^2+|z_1-z_3|^2=4$ if and only if $(z_1^2+z_2z_3)(z_2+z_3)=0$. Since we can always arrange for $z_1^2=-z_2z_3$ on the unit circle, $|z_2+z_3|$ can take any value between $0$ and $2$.

Hint: Writing $$fracbeta-gamma2=fracbeta-alpha2+fracalpha-gamma2$$ and combing this with trigonometric identity for the sum of cosine $$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$$ we get
beginaligncosleft(fracbeta-gamma2right)&=cosleft(fracbeta-alpha2+fracalpha-gamma2right)\
&=cosleft(fracbeta-alpha2right)cdotcosleft(fracalpha-gamma2right)-sinleft(fracbeta-alpha2right)cdotsinleft(fracalpha-gamma2right).endalign
Now rewrite the identity
$$sin^2left(fracalpha-beta2right)+sin^2left(fracalpha-gamma2right)=1$$
as
$$sin^2left(fracalpha-beta2right)=1-sin^2left(fracalpha-gamma2right)=cos^2left(fracalpha-gamma2right),$$
and plug this into the previous to conclude.