Let `z=x+iy` and `w=u+iv` be two complex numbers, such that `|z|=|w|=1` and `z^(2)+w^(2)=1.` Then, the number of ordered pairs (z, w) is equal to (where, `x, y, u, v in R and i^(2)=-1`)

Answer

To solve the problem, we need to analyze the given conditions for the complex numbers \( z \) and \( w \).

### Step 1: Understand the given conditions

We know that:
1. \( z = x + iy \) and \( w = u + iv \)
2. \( |z| = |w| = 1 \)
3. \( z^2 + w^2 = 1 \)

Since \( |z| = 1 \) and \( |w| = 1 \), we can express \( z \) and \( w \) in terms of trigonometric functions:
- Let \( z = e^{i\theta} = \cos(\theta) + i\sin(\theta) \)
- Let \( w = e^{i\phi} = \cos(\phi) + i\sin(\phi) \)

### Step 2: Substitute \( z \) and \( w \) into the equation

Now, substituting these into the equation \( z^2 + w^2 = 1 \):
\[ z^2 = e^{2i\theta} = \cos(2\theta) + i\sin(2\theta) \]
\[ w^2 = e^{2i\phi} = \cos(2\phi) + i\sin(2\phi) \]
Thus, we have:
\[ \cos(2\theta) + i\sin(2\theta) + \cos(2\phi) + i\sin(2\phi) = 1 \]

### Step 3: Separate real and imaginary parts

From the above equation, we can separate the real and imaginary parts:
1. Real part: \( \cos(2\theta) + \cos(2\phi) = 1 \)
2. Imaginary part: \( \sin(2\theta) + \sin(2\phi) = 0 \)

### Step 4: Analyze the equations

From the imaginary part:
\[ \sin(2\theta) + \sin(2\phi) = 0 \implies \sin(2\phi) = -\sin(2\theta) \]
This implies that:
\[ 2\phi = -2\theta + n\pi \quad \text{for some integer } n \]

From the real part:
\[ \cos(2\theta) + \cos(2\phi) = 1 \]
Using the identity for cosine:
\[ \cos(2\phi) = \cos(-2\theta + n\pi) = (-1)^n \cos(2\theta) \]
Thus, we have:
\[ \cos(2\theta) + (-1)^n \cos(2\theta) = 1 \]

### Step 5: Solve for \( n \)

This gives us two cases:
1. If \( n \) is even, \( (-1)^n = 1 \):
\[ 2\cos(2\theta) = 1 \implies \cos(2\theta) = \frac{1}{2} \implies 2\theta = \frac{\pi}{3} + 2k\pi \text{ or } 2\theta = \frac{5\pi}{3} + 2k\pi \]
Thus, \( \theta = \frac{\pi}{6} + k\pi \text{ or } \theta = \frac{5\pi}{6} + k\pi \).

2. If \( n \) is odd, \( (-1)^n = -1 \):
\[ 0 = 1 \quad \text{(not possible)} \]

### Step 6: Find corresponding \( \phi \)

For \( n \) even:
- For \( \theta = \frac{\pi}{6} \), \( \phi = \frac{5\pi}{6} \) or vice versa.
- For \( \theta = \frac{5\pi}{6} \), \( \phi = \frac{\pi}{6} \) or vice versa.

### Step 7: Count ordered pairs

Each \( \theta \) can correspond to two values of \( \phi \) (and vice versa). Thus, we have:
- \( (\theta_1, \phi_1) \)
- \( (\theta_1, \phi_2) \)
- \( (\theta_2, \phi_1) \)
- \( (\theta_2, \phi_2) \)

This gives us a total of 4 ordered pairs \( (z, w) \).

### Final Answer
The number of ordered pairs \( (z, w) \) is \( \boxed{4} \).