If `Re(z)<0` then the value of `(1+z+z^2+.....+z^n)` cannot exceed

Options

A. `|z^(n)|-1/|z|`
B. `n|z|^(n)+1`
C. `|z|^(n)-1/|z|`
D. `|z|^(n)+1/|z|`

Answer

To solve the problem, we need to evaluate the expression \( |1 + z + z^2 + \ldots + z^n| \) given the condition that \( \text{Re}(z) < 0 \).

### Step-by-Step Solution:

1. **Identify the Series**: The expression \( 1 + z + z^2 + \ldots + z^n \) is a geometric series. The formula for the sum of a geometric series is:
\[ S_n = \frac{a(1 - r^{n+1})}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = 1 \) and \( r = z \).

2. **Apply the Formula**: Using the formula for the sum of the geometric series, we have:
\[ S_n = \frac{1 - z^{n+1}}{1 - z} \]

3. **Take the Modulus**: We need to find the modulus of this sum:
\[ |S_n| = \left| \frac{1 - z^{n+1}}{1 - z} \right| = \frac{|1 - z^{n+1}|}{|1 - z|} \]

4. **Analyze the Denominator**: Since \( \text{Re}(z) < 0 \), we can deduce that \( |1 - z| \) is greater than \( |z| \). This is because the point \( z \) lies in the left half of the complex plane, making the distance from \( 1 \) (which is on the real axis) greater than the distance from the origin.

5. **Use the Relationship**: From the previous step, we can establish that:
\[ |1 - z| > |z| \]

6. **Substitute Back**: Now we can express \( |S_n| \) in terms of \( |z| \):
\[ |S_n| = \frac{|1 - z^{n+1}|}{|1 - z|} < \frac{|1 - z^{n+1}|}{|z|} \]

7. **Estimate the Numerator**: Since \( |z^{n+1}| = |z|^{n+1} \), we can analyze \( |1 - z^{n+1}| \). For large \( n \), \( |z^{n+1}| \) becomes significant, and thus:
\[ |1 - z^{n+1}| \approx |z|^{n+1} \]

8. **Final Expression**: Therefore, we can conclude:
\[ |S_n| < \frac{|z|^{n+1}}{|z|} = |z|^n \]

### Conclusion:
Thus, the value of \( |1 + z + z^2 + \ldots + z^n| \) cannot exceed \( |z|^n \).