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 \( |z + 16| = 4 |z + 1| \) for the complex number \( z \), we can 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 us:
\[ |z + 16|^2 = (4 |z + 1|)^2 \]
This simplifies to:
\[ |z + 16|^2 = 16 |z + 1|^2 \]

### Step 2: Expand both sides
Using the property of modulus, we can express the left side as:
\[ (z + 16)(\overline{z + 16}) = (z + 16)(\overline{z} + 16) = z\overline{z} + 16z + 16\overline{z} + 256 \]
And the right side as:
\[ 16 |z + 1|^2 = 16(z + 1)(\overline{z + 1}) = 16(z + 1)(\overline{z} + 1) = 16(z\overline{z} + z + \overline{z} + 1) \]

### Step 3: Set the expanded forms equal
Now we have:
\[ z\overline{z} + 16z + 16\overline{z} + 256 = 16(z\overline{z} + z + \overline{z} + 1) \]

### Step 4: Distribute on the right side
Expanding the right side gives:
\[ 16z\overline{z} + 16z + 16\overline{z} + 16 \]

### Step 5: Rearrange the equation
Now, we can rearrange the equation:
\[ z\overline{z} + 16z + 16\overline{z} + 256 = 16z\overline{z} + 16z + 16\overline{z} + 16 \]
Subtract \( 16z + 16\overline{z} \) from both sides:
\[ z\overline{z} + 256 = 16z\overline{z} + 16 \]

### Step 6: Move all terms involving \( z\overline{z} \) to one side
Rearranging gives:
\[ z\overline{z} - 16z\overline{z} = 16 - 256 \]
This simplifies to:
\[ -15z\overline{z} = -240 \]

### Step 7: Solve for \( z\overline{z} \)
Dividing both sides by -15 gives:
\[ z\overline{z} = \frac{240}{15} = 16 \]

### Step 8: Find \( |z|^2 \)
Since \( z\overline{z} = |z|^2 \), we have:
\[ |z|^2 = 16 \]

### Conclusion
Thus, the value of \( |z|^2 \) is \( 16 \).

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