If `z=a+ib` satisfies `"arg"(z-1)="arg"(z+3i)`, then `(a-1):b=`

Options

A. `2:1`
B. `1:3`
C. `-1:3`
D. none of these

Explanation

To solve the problem, we need to analyze the condition given by the equality of arguments. The problem states that if \( z = a + ib \), then \( \arg(z - 1) = \arg(z + 3i) \).

### Step-by-Step Solution:

1. **Express the complex numbers**:
- We have \( z = a + ib \).
- Therefore, \( z - 1 = (a - 1) + ib \) and \( z + 3i = a + i(b + 3) \).

2. **Set up the argument equality**:
- According to the problem, we have:
\[ \arg(z - 1) = \arg(z + 3i) \]
- This translates to:
\[ \arg((a - 1) + ib) = \arg(a + i(b + 3)) \]

3. **Use the definition of argument**:
- The argument of a complex number \( x + iy \) is given by \( \tan^{-1}(\frac{y}{x}) \).
- Therefore:
\[ \arg((a - 1) + ib) = \tan^{-1}\left(\frac{b}{a - 1}\right) \]
\[ \arg(a + i(b + 3)) = \tan^{-1}\left(\frac{b + 3}{a}\right) \]

4. **Set the arguments equal**:
- We set the two expressions equal:
\[ \tan^{-1}\left(\frac{b}{a - 1}\right) = \tan^{-1}\left(\frac{b + 3}{a}\right) \]

5. **Eliminate the tangent function**:
- Since the tangent function is one-to-one in the range of \( \tan^{-1} \), we can equate the arguments:
\[ \frac{b}{a - 1} = \frac{b + 3}{a} \]

6. **Cross-multiply to eliminate the fractions**:
- Cross-multiplying gives:
\[ b \cdot a = (b + 3)(a - 1) \]

7. **Expand the right-hand side**:
- Expanding the right-hand side:
\[ ba = ba - b + 3a - 3 \]

8. **Simplify the equation**:
- Cancel \( ba \) from both sides:
\[ 0 = -b + 3a - 3 \]
- Rearranging gives:
\[ 3a - b - 3 = 0 \]

9. **Express \( b \) in terms of \( a \)**:
- Rearranging gives:
\[ b = 3a - 3 \]

10. **Find the ratio \( (a - 1) : b \)**:
- Substitute \( b \) into the ratio:
\[ a - 1 : b = a - 1 : (3a - 3) \]
- This can be simplified:
\[ = a - 1 : 3(a - 1) \]
- Thus, the ratio simplifies to:
\[ (a - 1) : b = 1 : 3 \]

### Final Answer:
\[ (a - 1) : b = 1 : 3 \]