If P and Q are represented by the complex numbers `z_(1)` and `z_(2)` such that `|1/z_(2)+1/z_(1)|=|1/z_(2)-1/z_(1)|`, then

Options

A. (a) `DeltaOPQ` (where O is the origin) is equilateral.
B. (b) `DeltaOPQ` is right angled
C. (c) the circumcentre of `DeltaOPQ` is `(1)/(2)(z_(1) + z_(2))`
D. (d) the circumcentre of `DeltaOPQ` is `(1)/(2)(z_(1)-z_(2))`

Explanation

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

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

### Step 1: Rewrite the equation
We can rewrite the left-hand side and right-hand side using a common denominator:

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

### Step 2: Remove the denominator
Since \( z_1 \) and \( z_2 \) are not zero, we can multiply both sides by \( |z_1 z_2| \):

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

### Step 3: Square both sides
Now, squaring both sides gives us:

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

### Step 4: Expand both sides
Using the property of modulus, we expand both sides:

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

This leads to:

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

### Step 5: Simplify the equation
Cancelling \( |z_1|^2 + |z_2|^2 \) from both sides gives:

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

This simplifies to:

\[ 2(z_1 \overline{z_2} + z_2 \overline{z_1}) = 0 \]

### Step 6: Rearranging
Thus, we have:

\[ z_1 \overline{z_2} + z_2 \overline{z_1} = 0 \]

### Step 7: Divide by \( z_2 \overline{z_2} \)
Dividing both sides by \( z_2 \overline{z_2} \) gives:

\[ \frac{z_1}{z_2} + \frac{z_2}{z_1} = 0 \]

Let \( z = \frac{z_1}{z_2} \). Then we have:

\[ z + \frac{1}{z} = 0 \]

### Step 8: Solve for \( z \)
Multiplying through by \( z \) gives:

\[ z^2 + 1 = 0 \]

Thus,

\[ z^2 = -1 \implies z = i \text{ or } z = -i \]

### Step 9: Conclusion about angles
This implies that:

\[ \frac{z_1}{z_2} = i \text{ or } \frac{z_1}{z_2} = -i \]

This means that the argument of \( \frac{z_1}{z_2} \) is \( \frac{\pi}{2} \) or \( -\frac{\pi}{2} \), indicating that \( z_1 \) and \( z_2 \) are perpendicular to each other.

### Final Result
Thus, the angle between the vectors represented by \( z_1 \) and \( z_2 \) is \( \frac{\pi}{2} \), confirming that they form a right triangle.