Animated Solution for Mathematics - Complex Numbers: If $z_1,z_2$ and $z_3$ are complex numbers such that $|z_1|=|z_2|=|z_3|=|1/z_1+1/z_2+1/z_3|=1$, then $|z_1+z_2+z_3|$ is

Welcome, future engineer! Today, we are going to unravel a problem that looks intimidating at first glance but hides a profound elegance. We are dealing with complex numbers $z_1$, $z_2$, and $z_3$ that are constrained to the unit circle.

When you see the condition $|z_1|=|z_2|=|z_3|=1$, I want you to stop and visualize the Argand plane. Imagine a circle of radius $1$ centered at the origin.

These three complex numbers are not just random points; they are vectors anchored at the origin, stretching out to touch the boundary of this circle. This is our starting point: a geometric playground where the magnitude of every vector is exactly unity.

The problem then introduces a second condition:

At first, this looks messy. We are dealing with reciprocals, and adding complex numbers in the denominator is usually a recipe for algebraic disaster. But wait! We have a secret weapon.

We know that for any complex number $z$, the square of its modulus is given by the product of the number and its conjugate: $|z{|}^2=z\cdot \bar{z}$. Since we know $|z_1|=1$, it follows that $|z_1{|}^2=1$, which implies $z_1\cdot {\bar{z}}_1=1$.

If we rearrange this, we get the most beautiful substitution in complex algebra:

Suddenly, the reciprocals are gone! We have replaced them with conjugates. This is the moment where the problem shifts from a difficult calculation to a simple property check.

Now, let us apply this to our given equation. We substitute $\frac{1}{z_1}$ with ${\bar{z}}_1$, $\frac{1}{z_2}$ with ${\bar{z}}_2$, and $\frac{1}{z_3}$ with ${\bar{z}}_3$. Our equation transforms into:

We are almost there. We have the sum of conjugates, but we want the modulus of the sum of the original numbers. Is there a relationship between the sum of conjugates and the conjugate of a sum?

Absolutely. The conjugate of a sum is equal to the sum of the conjugates:

This allows us to rewrite our expression as $|\overline{z_1+z_2+z_3}|=1$.

We are at the finish line. We have the modulus of a conjugate, $|\overline{z_1+z_2+z_3}|$. Geometrically, taking the conjugate is just a reflection across the real axis.

Does reflecting a vector change its length? Of course not! The distance from the origin remains invariant. Therefore, the modulus of a complex number is identical to the modulus of its conjugate: $|\bar{Z}|=|Z|$.

Applying this final property, we conclude that:

It is a stunning result. Despite the complexity of the initial setup, the symmetry of the unit circle guides us to a clean, simple answer. Never be afraid of the algebra; look for the geometric symmetry hidden underneath, and the path will always reveal itself.