The distance of the roots of the equation `tan theta_(0) z^(n) + tan theta_(1) z^(n-1) + …+ tan theta_(n) = 3` from `z=0`, where `theta_(0) , theta_(1) , theta_(2),…, theta_(n) in [0, (pi)/(4)]` satisfy

Options

A. greater than 2/3
B. less than 2/3
C. greater than `| cos theta_(1)|+|cos theta_(2)|+…+|cos theta_(n)|`
D. less than `| cos theta_(1)|+|cos theta_(2)|+…+|cos theta_(n)|`

Answer

To solve the equation \( \tan \theta_0 z^n + \tan \theta_1 z^{n-1} + \ldots + \tan \theta_n = 3 \) and find the distance of the roots from \( z = 0 \), we will follow these steps:

### Step 1: Understand the Equation
The given equation is a polynomial in \( z \) with coefficients \( \tan \theta_k \) where \( \theta_k \) varies from \( 0 \) to \( \frac{\pi}{4} \). We need to analyze the roots of this polynomial.

### Step 2: Use the Triangle Inequality
Using the triangle inequality for complex numbers, we know that:
\[ |z_1 + z_2 + \ldots + z_n| \leq |z_1| + |z_2| + \ldots + |z_n| \]
Applying this to our polynomial, we can write:
\[ |\tan \theta_0 z^n + \tan \theta_1 z^{n-1} + \ldots + \tan \theta_n| \leq |\tan \theta_0| |z^n| + |\tan \theta_1| |z^{n-1}| + \ldots + |\tan \theta_n| \]

### Step 3: Analyze the Coefficients
Since \( \theta_k \in [0, \frac{\pi}{4}] \), we have:
\[ 0 \leq \tan \theta_k \leq 1 \]
Thus, the coefficients \( \tan \theta_k \) are bounded between \( 0 \) and \( 1 \).

### Step 4: Set Up the Inequality
From the equation, we have:
\[ |\tan \theta_0 z^n + \tan \theta_1 z^{n-1} + \ldots + \tan \theta_n| = 3 \]
So we can write:
\[ 3 \leq |\tan \theta_0| |z^n| + |\tan \theta_1| |z^{n-1}| + \ldots + |\tan \theta_n| \]

### Step 5: Substitute Maximum Values
Substituting the maximum values of \( \tan \theta_k \):
\[ 3 \leq |z^n| + |z^{n-1}| + \ldots + |z| + 1 \]
This is a geometric series with \( n+1 \) terms. The sum can be expressed as:
\[ |z|^n + |z|^{n-1} + \ldots + |z| + 1 = \frac{|z|^{n+1} - 1}{|z| - 1} \quad \text{(for } |z| \neq 1\text{)} \]

### Step 6: Analyze the Limit as \( n \to \infty \)
As \( n \) approaches infinity, we consider the case when \( |z| < 1 \):
\[ 3 \leq \frac{|z|^{n+1} - 1}{|z| - 1} \]
This implies that for large \( n \), the polynomial's roots must be such that:
\[ |z| > \frac{2}{3} \]

### Step 7: Conclusion
Thus, the distance of the roots from \( z = 0 \) must satisfy:
\[ |z| > \frac{2}{3} \]