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 will follow these steps:

### Step 1: Represent the complex numbers
Let \( z = x + iy \) and \( w = u + iv \). Given that \( |z| = |w| = 1 \), we can express \( z \) and \( w \) in exponential form:
\[ z = e^{i\alpha} \quad \text{and} \quad w = e^{i\beta} \]
where \( \alpha \) and \( \beta \) are angles in radians.

### Step 2: Use the given condition
We are given that:
\[ z^2 + w^2 = 1 \]
Substituting the exponential forms:
\[ e^{2i\alpha} + e^{2i\beta} = 1 \]

### Step 3: Rewrite in terms of cosine and sine
Using Euler's formula, we can rewrite the equation:
\[ \cos(2\alpha) + i\sin(2\alpha) + \cos(2\beta) + i\sin(2\beta) = 1 \]
This implies that the real part must equal 1 and the imaginary part must equal 0:
1. \(\cos(2\alpha) + \cos(2\beta) = 1\)
2. \(\sin(2\alpha) + \sin(2\beta) = 0\)

### Step 4: Analyze the equations
From the second equation, we can express:
\[ \sin(2\beta) = -\sin(2\alpha) \]
Using the sine addition formula, we can write:
\[ \sin(2\beta) = \sin(2\alpha + \pi) \quad \text{(since } \sin(x + \pi) = -\sin x\text{)} \]
This implies:
\[ 2\beta = 2\alpha + \pi + 2k\pi \quad \text{for some integer } k \]
Thus, we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = \alpha + \frac{\pi}{2} + k\pi \]

### Step 5: Substitute back into the cosine equation
Now substituting \( \beta \) into the cosine equation:
\[ \cos(2\alpha) + \cos(2(\alpha + \frac{\pi}{2} + k\pi)) = 1 \]
Since \( \cos(2(\alpha + \frac{\pi}{2})) = -\sin(2\alpha) \) (because of the phase shift), we have:
\[ \cos(2\alpha) - \sin(2\alpha) = 1 \]

### Step 6: Solve for \( \alpha \)
Rearranging gives:
\[ \cos(2\alpha) - 1 = \sin(2\alpha) \]
This can be solved using trigonometric identities.

### Step 7: Count the solutions
We find that the solutions for \( \alpha \) in the range \( [0, 2\pi) \) yield specific angles. Each valid \( \alpha \) corresponds to a unique \( \beta \), and we can find that there are 8 ordered pairs \( (z, w) \) that satisfy the conditions.

### Final Answer
Thus, the number of ordered pairs \( (z, w) \) is:
\[ \boxed{8} \]