If for the complex numbers `z_1` and `z_2`, `|z_1+z_2|=|z_1-z_2|`, then `Arg(z_1)-Arg(z_2)` is equal to

Options

A. `pi`
B. `pi//2`
C. `pi//4`
D. 4

Answer

To solve the problem, we need to find the value of \( \text{Arg}(z_1) - \text{Arg}(z_2) \) given that \( |z_1 + z_2| = |z_1 - z_2| \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We start with the condition \( |z_1 + z_2| = |z_1 - z_2| \). This means that the distance from the origin to the point representing \( z_1 + z_2 \) is equal to the distance to the point representing \( z_1 - z_2 \).

2. **Expressing Complex Numbers**:
Let \( z_1 = a + ib \) and \( z_2 = x + iy \), where \( a, b, x, y \) are real numbers.

3. **Applying the Modulus Condition**:
The modulus condition can be expressed as:
\[ |(a + x) + i(b + y)| = |(a - x) + i(b - y)| \]
This translates to:
\[ \sqrt{(a + x)^2 + (b + y)^2} = \sqrt{(a - x)^2 + (b - y)^2} \]

4. **Squaring Both Sides**:
Squaring both sides gives:
\[ (a + x)^2 + (b + y)^2 = (a - x)^2 + (b - y)^2 \]

5. **Expanding Both Sides**:
Expanding both sides results in:
\[ a^2 + 2ax + x^2 + b^2 + 2by + y^2 = a^2 - 2ax + x^2 + b^2 - 2by + y^2 \]

6. **Simplifying the Equation**:
Canceling out common terms, we have:
\[ 2ax + 2by = -2ax - 2by \]
This simplifies to:
\[ 4ax + 4by = 0 \quad \Rightarrow \quad ax + by = 0 \]

7. **Finding the Ratio**:
From \( ax + by = 0 \), we can express this as:
\[ \frac{a}{b} = -\frac{y}{x} \]

8. **Finding Arguments**:
The argument of a complex number \( z \) can be expressed as:
\[ \text{Arg}(z_1) = \tan^{-1}\left(\frac{b}{a}\right), \quad \text{Arg}(z_2) = \tan^{-1}\left(\frac{y}{x}\right) \]

9. **Substituting the Ratio**:
We substitute \( \frac{y}{x} = -\frac{a}{b} \) into the expression for \( \text{Arg}(z_2) \):
\[ \text{Arg}(z_2) = \tan^{-1}\left(-\frac{a}{b}\right) \]

10. **Calculating the Difference**:
Now we find:
\[ \text{Arg}(z_1) - \text{Arg}(z_2) = \tan^{-1}\left(\frac{b}{a}\right) - \tan^{-1}\left(-\frac{a}{b}\right) \]

11. **Using the Tangent Addition Formula**:
Using the formula \( \tan^{-1}(u) - \tan^{-1}(v) = \tan^{-1}\left(\frac{u - v}{1 + uv}\right) \):
\[ \text{Arg}(z_1) - \text{Arg}(z_2) = \tan^{-1}\left(\frac{\frac{b}{a} + \frac{a}{b}}{1 - \frac{b}{a} \cdot \frac{a}{b}}\right) \]

12. **Simplifying the Expression**:
This simplifies to:
\[ = \tan^{-1}\left(\frac{\frac{b^2 + a^2}{ab}}{1 - 1}\right) = \tan^{-1}(\infty) = \frac{\pi}{2} \]

### Final Result:
Thus, we conclude that:
\[ \text{Arg}(z_1) - \text{Arg}(z_2) = \frac{\pi}{2} \]