Given the complex numbers \(z_1\), \(z_2\), and \(z_3\), we know that:
Since \(|z_1| = |z_2| = |z_3| = 1\), each of these complex numbers lies on the unit circle in the complex plane. So, we can express them as \(z_1 = e^{i\theta_1}\), \(z_2 = e^{i\theta_2}\), and \(z_3 = e^{i\theta_3}\) for some real angles \(\theta_1\), \(\theta_2\), and \(\theta_3\).
The expression \(\frac{1}{z_1}\) can be represented as the conjugate for unit complex numbers, i.e., \(\frac{1}{z_1} = \overline{z_1} = e^{-i\theta_1}\).
Similarly, \(\frac{1}{z_2} = \overline{z_2} = e^{-i\theta_2}\) and \(\frac{1}{z_3} = \overline{z_3} = e^{-i\theta_3}\).
Thus, \(\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} = e^{-i\theta_1} + e^{-i\theta_2} + e^{-i\theta_3}\), and we know:
\(\left| e^{-i\theta_1} + e^{-i\theta_2} + e^{-i\theta_3} \right| = 1\)
This implies these vectors in the complex plane add up to a unit magnitude. For example, if they are all equal angles apart, they could form an equilateral triangle centered at the origin.
Now, examine \(|z_1 + z_2 + z_3|\):
Using the relationship between a complex number and its conjugate: \(z_1 = e^{i\theta_1}\), \(z_2 = e^{i\theta_2}\), and \(z_3 = e^{i\theta_3}\), and the fact that the magnitudes are all 1, we establish that:
Thus, by symmetry and calculating such arrangements, we find:
\(|z_1 + z_2 + z_3|^2 = |z_1|^2 + |z_2|^2 + |z_3|^2 \pm 2 \times Re(z_1z_2 + z_2z_3 + z_3z_1) = 3 \pm 2 \cdot 0 = 1\)
Hence, \(|z_1 + z_2 + z_3| = 1\).
Therefore, the correct answer is: equal to 1.