Let `z` be a complex number satisfying `|z+16|=4|z+1|`. Then

Options

A. `|z|=4`
B. `|z|=5`
C. `|z|=6`
D. `3 lt |z| lt68`

Explanation

To solve the problem \( |z + 16| = 4 |z + 1| \), we will follow these steps:

### Step 1: Square both sides of the equation
We start with the equation:
\[ |z + 16| = 4 |z + 1| \]
Squaring both sides gives:
\[ |z + 16|^2 = (4 |z + 1|)^2 \]
This simplifies to:
\[ |z + 16|^2 = 16 |z + 1|^2 \]

### Step 2: Express the moduli in terms of \( z \)
Recall that for any complex number \( z \), \( |z|^2 = z \cdot \overline{z} \). Thus, we can express \( |z + 16|^2 \) and \( |z + 1|^2 \):
\[ |z + 16|^2 = (z + 16)(\overline{z} + 16) = z\overline{z} + 16z + 16\overline{z} + 256 \]
\[ |z + 1|^2 = (z + 1)(\overline{z} + 1) = z\overline{z} + z + \overline{z} + 1 \]

### Step 3: Substitute back into the equation
Substituting these expressions into the squared equation gives:
\[ z\overline{z} + 16z + 16\overline{z} + 256 = 16(z\overline{z} + z + \overline{z} + 1) \]

### Step 4: Expand the right-hand side
Expanding the right-hand side:
\[ z\overline{z} + 16z + 16\overline{z} + 256 = 16z\overline{z} + 16z + 16\overline{z} + 16 \]

### Step 5: Rearrange the equation
Now, we can rearrange the equation:
\[ z\overline{z} + 16z + 16\overline{z} + 256 - 16z\overline{z} - 16z - 16\overline{z} - 16 = 0 \]
This simplifies to:
\[ (1 - 16)z\overline{z} + (256 - 16) = 0 \]
\[ -15z\overline{z} + 240 = 0 \]

### Step 6: Solve for \( z\overline{z} \)
Now, we can solve for \( z\overline{z} \):
\[ -15z\overline{z} = -240 \]
\[ z\overline{z} = \frac{240}{15} = 16 \]

### Step 7: Conclusion
Since \( z\overline{z} = |z|^2 \), we have:
\[ |z|^2 = 16 \]
Thus, the modulus of \( z \) is:
\[ |z| = 4 \]

### Final Answer
The solution indicates that the modulus of the complex number \( z \) is \( 4 \).

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