Suppose two complex numbers `z=a+ib`, `w=c+id` satisfy the equation `(z+w)/(z)=(w)/(z+w)`. Then

Options

A. both `a` and `c` are zeros
B. both `b` and `d` are zeros
C. both `b` and `d` must be non zeros
D. at least one of `b` and `d` is non zero

Explanation

To solve the equation \(\frac{z+w}{z} = \frac{w}{z+w}\) where \(z = a + ib\) and \(w = c + id\), we will follow these steps:

### Step 1: Cross Multiply
Starting with the equation:
\[ \frac{z+w}{z} = \frac{w}{z+w} \]
Cross multiplying gives:
\[ (z + w)(z + w) = zw \]

### Step 2: Expand the Left Side
Expanding the left side:
\[ (z + w)^2 = zw \]
This expands to:
\[ z^2 + 2zw + w^2 = zw \]

### Step 3: Rearranging the Equation
Rearranging gives:
\[ z^2 + w^2 + zw - zw = 0 \]
This simplifies to:
\[ z^2 + w^2 + zw = 0 \]

### Step 4: Divide by \(w^2\)
Now, divide the entire equation by \(w^2\):
\[ \frac{z^2}{w^2} + 1 + \frac{z}{w} = 0 \]
Let \(t = \frac{z}{w}\). Then, we have:
\[ t^2 + 1 + t = 0 \]

### Step 5: Forming a Quadratic Equation
This can be rearranged to form a quadratic equation:
\[ t^2 + t + 1 = 0 \]

### Step 6: Finding the Roots
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = 1\), and \(c = 1\):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ t = \frac{-1 \pm i\sqrt{3}}{2} \]

### Step 7: Interpreting the Results
The values of \(t\) correspond to the complex numbers:
\[ t = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad t = \frac{-1 - i\sqrt{3}}{2} \]
These represent points on the complex plane with arguments of \(120^\circ\) and \(240^\circ\).

### Step 8: Conclusion about \(b\) and \(d\)
From the results, we can conclude:
- Both \(b\) and \(d\) cannot be zero.
- At least one of \(b\) or \(d\) must be non-zero for the equation to hold.