If `|z_1|=|z_2|=|z_3|=......=|z_n|=1`, then `|z_1+z_2+z_3+......+z_n|=`

Options

A. `n`
B. `|(1)/(z)+(1)/(z_(2))+...+(1)/(z_(n))|`
C. 0
D. None of these

Answer

To solve the problem, we need to find the modulus of the sum of complex numbers \( z_1, z_2, z_3, \ldots, z_n \) given that each of them has a modulus of 1, i.e., \( |z_1| = |z_2| = |z_3| = \ldots = |z_n| = 1 \).

### Step-by-step Solution:

1. **Understanding the Modulus Condition**:
Since \( |z_k| = 1 \) for each \( k \), we can express each complex number \( z_k \) in the form:
\[ z_k = e^{i\theta_k} \]
where \( \theta_k \) is the argument of \( z_k \).

2. **Expressing the Sum**:
We need to find the modulus of the sum:
\[ |z_1 + z_2 + z_3 + \ldots + z_n| \]

3. **Using Properties of Conjugates**:
The conjugate of the sum can be expressed as:
\[ \overline{z_1 + z_2 + z_3 + \ldots + z_n} = \overline{z_1} + \overline{z_2} + \overline{z_3} + \ldots + \overline{z_n} \]
Since \( |z_k| = 1 \), we have \( \overline{z_k} = \frac{1}{z_k} \). Thus, we can write:
\[ \overline{z_1 + z_2 + z_3 + \ldots + z_n} = \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \ldots + \frac{1}{z_n} \]

4. **Finding the Modulus**:
Now, we can use the property of modulus:
\[ |z_1 + z_2 + z_3 + \ldots + z_n| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \ldots + \frac{1}{z_n} \right| \]
Since \( |z_k| = 1 \), we know that \( \left| \frac{1}{z_k} \right| = 1 \).

5. **Conclusion**:
Therefore, the modulus of the sum of the complex numbers can be expressed as:
\[ |z_1 + z_2 + z_3 + \ldots + z_n| = |z_1| + |z_2| + |z_3| + \ldots + |z_n| = n \]
However, this is only true if all \( z_k \) are aligned in the same direction (i.e., they are all equal). In general, the modulus can vary depending on the angles \( \theta_k \).

Thus, the final answer is:
\[ |z_1 + z_2 + z_3 + \ldots + z_n| \leq n \]