If `z_(1)` and `z_(2)` are two complex numbers satisying the equation.
`|(iz_(1)+z_(2))/(iz_(1)-z_(2))|=1`, then `z_(1)/z_(2)` is

Options

A. 0
B. purely real
C. negative real
D. purely imaginary

Explanation

To solve the given problem, we start with the equation:

\[ \left| \frac{i z_1 + z_2}{i z_1 - z_2} \right| = 1 \]

### Step 1: Understanding the Modulus Condition

The condition \(\left| \frac{a}{b} \right| = 1\) implies that \(|a| = |b|\). Therefore, we can rewrite our equation as:

\[ |i z_1 + z_2| = |i z_1 - z_2| \]

### Step 2: Squaring Both Sides

To eliminate the modulus, we can square both sides:

\[ |i z_1 + z_2|^2 = |i z_1 - z_2|^2 \]

### Step 3: Expanding the Modulus Squared

Using the property that \(|z|^2 = z \cdot \overline{z}\), we expand both sides:

\[ (i z_1 + z_2)(-i \overline{z_1} + \overline{z_2}) = (i z_1 - z_2)(-i \overline{z_1} - \overline{z_2}) \]

Calculating the left-hand side:

\[ |i z_1 + z_2|^2 = (i z_1 + z_2)(-i \overline{z_1} + \overline{z_2}) = z_1 \overline{z_1} + z_2 \overline{z_2} + i z_1 \overline{z_2} - i z_2 \overline{z_1} \]

Calculating the right-hand side:

\[ |i z_1 - z_2|^2 = (i z_1 - z_2)(-i \overline{z_1} - \overline{z_2}) = z_1 \overline{z_1} + z_2 \overline{z_2} - i z_1 \overline{z_2} + i z_2 \overline{z_1} \]

### Step 4: Setting the Two Sides Equal

Now we have:

\[ z_1 \overline{z_1} + z_2 \overline{z_2} + i z_1 \overline{z_2} - i z_2 \overline{z_1} = z_1 \overline{z_1} + z_2 \overline{z_2} - i z_1 \overline{z_2} + i z_2 \overline{z_1} \]

### Step 5: Simplifying the Equation

Subtract \(z_1 \overline{z_1} + z_2 \overline{z_2}\) from both sides:

\[ i z_1 \overline{z_2} - i z_2 \overline{z_1} = -i z_1 \overline{z_2} + i z_2 \overline{z_1} \]

Combining like terms gives us:

\[ 2i z_1 \overline{z_2} = 2i z_2 \overline{z_1} \]

### Step 6: Dividing Both Sides by \(2i\)

Assuming \(i \neq 0\), we can divide both sides by \(2i\):

\[ z_1 \overline{z_2} = z_2 \overline{z_1} \]

### Step 7: Rearranging the Equation

Rearranging gives us:

\[ \frac{z_1}{z_2} = \frac{\overline{z_1}}{\overline{z_2}} \]

### Step 8: Conclusion

This implies that \(\frac{z_1}{z_2}\) is a real number. Therefore, we conclude that:

\[ \frac{z_1}{z_2} \text{ is real.} \]