If ` |z + omega|^(2) = |z|^(2)+|omega|^(2)`, where z and ` omega ` are complex numbers , then

Options

A. `z/omega ` is purely real
B. `z/omega` is purely imaginary
C. `zbaromega + barz omega =0`
D. ` amp ( z/omega) = pi/2`

Answer

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

\[ |z + \omega|^2 = |z|^2 + |\omega|^2 \]

where \( z \) and \( \omega \) are complex numbers.

### Step 1: Express \( z \) and \( \omega \)
Let \( z = x + iy \) where \( x \) and \( y \) are real numbers. We are also given that \( \omega = \frac{1}{2} + i \frac{\sqrt{3}}{2} \).

### Step 2: Calculate \( |z|^2 \) and \( |\omega|^2 \)
The modulus squared of \( z \) is calculated as:

\[ |z|^2 = x^2 + y^2 \]

For \( \omega \):

\[ |\omega|^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1 \]

### Step 3: Substitute into the equation
Now substitute these values back into the original equation:

\[ |z + \omega|^2 = |z|^2 + |\omega|^2 \implies |z + \omega|^2 = x^2 + y^2 + 1 \]

### Step 4: Calculate \( |z + \omega|^2 \)
Now calculate \( z + \omega \):

\[ z + \omega = \left(x + \frac{1}{2}\right) + i\left(y + \frac{\sqrt{3}}{2}\right) \]

Now, we find the modulus squared:

\[ |z + \omega|^2 = \left(x + \frac{1}{2}\right)^2 + \left(y + \frac{\sqrt{3}}{2}\right)^2 \]

### Step 5: Expand and set the equation
Expanding this gives:

\[ \left(x + \frac{1}{2}\right)^2 = x^2 + x + \frac{1}{4} \]
\[ \left(y + \frac{\sqrt{3}}{2}\right)^2 = y^2 + y\sqrt{3} + \frac{3}{4} \]

Combining these, we have:

\[ |z + \omega|^2 = x^2 + y^2 + x + y\sqrt{3} + 1 \]

### Step 6: Set the two expressions equal
Now we set the two expressions for \( |z + \omega|^2 \) equal to each other:

\[ x^2 + y^2 + x + y\sqrt{3} + 1 = x^2 + y^2 + 1 \]

### Step 7: Simplify
Subtract \( x^2 + y^2 + 1 \) from both sides:

\[ x + y\sqrt{3} = 0 \]

### Step 8: Rearranging
From this, we can express:

\[ x = -y\sqrt{3} \]

### Step 9: Find the ratio \( \frac{z}{\omega} \)
Now, we need to find \( \frac{z}{\omega} \):

\[ \frac{z}{\omega} = \frac{x + iy}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} = \frac{2(x + iy)}{1 + i\sqrt{3}} \]

### Step 10: Rationalize the denominator
To rationalize:

\[ \frac{2(x + iy)(1 - i\sqrt{3})}{(1 + i\sqrt{3})(1 - i\sqrt{3})} = \frac{2((x + iy)(1 - i\sqrt{3}))}{1 + 3} = \frac{2((x + iy)(1 - i\sqrt{3}))}{4} \]

### Step 11: Simplify
This simplifies to:

\[ \frac{(x + iy)(1 - i\sqrt{3})}{2} \]

### Step 12: Substitute \( x = -y\sqrt{3} \)
Substituting \( x = -y\sqrt{3} \):

\[ \frac{(-y\sqrt{3} + iy)(1 - i\sqrt{3})}{2} \]

### Conclusion
This shows that \( \frac{z}{\omega} \) is purely imaginary, confirming that the argument \( \frac{z}{\omega} \) is \( \frac{\pi}{2} \).