The value of `z` satisfying the equation `logz+logz^2+dot+logz^n=0i s`

Options

A. `cos.(4mpi)/(n(n+1)) +i sin.(4mpi)/(n(n+1)),m = 0,1,2,...`
B. `cos.(4mpi)/(n(n+1)) -i sin.(4mpi)/(n(n+1)),m = 0,1,2,...`
C. `sin.(4mpi)/(n) +i cos.(4mpi)/(n),m = 0,1,2,...`
D. 0

Explanation

To find the value of \( z \) satisfying the equation \( \log z + \log z^2 + \ldots + \log z^n = 0 \), we can follow these steps:

### Step 1: Combine the logarithms
Using the property of logarithms that states \( \log a + \log b = \log(ab) \), we can combine the terms:
\[ \log z + \log z^2 + \log z^3 + \ldots + \log z^n = \log(z \cdot z^2 \cdot z^3 \cdots z^n) \]
This simplifies to:
\[ \log(z^{1 + 2 + 3 + \ldots + n}) = 0 \]

### Step 2: Calculate the sum of the series
The sum of the first \( n \) natural numbers is given by the formula:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Thus, we can rewrite the logarithmic equation as:
\[ \log(z^{\frac{n(n + 1)}{2}}) = 0 \]

### Step 3: Exponentiate both sides
Exponentiating both sides gives us:
\[ z^{\frac{n(n + 1)}{2}} = 1 \]

### Step 4: Solve for \( z \)
The equation \( z^{\frac{n(n + 1)}{2}} = 1 \) implies that \( z \) can be expressed in terms of complex exponentials:
\[ z = e^{i \frac{2m\pi}{\frac{n(n + 1)}{2}}} \quad \text{for } m = 0, 1, 2, \ldots, \frac{n(n + 1)}{2} - 1 \]
This simplifies to:
\[ z = e^{i \frac{4m\pi}{n(n + 1)}} \]

### Final Result
Thus, the value of \( z \) satisfying the original equation is:
\[ z = e^{i \frac{4m\pi}{n(n + 1)}} \]
where \( m \) is an integer ranging from \( 0 \) to \( \frac{n(n + 1)}{2} - 1 \).

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