13.	a)	Solve (x^2+y^2)(p^2+q^2) = 1.
	b)	Solve the equation   zpq=p+q.

Let's solve the given equations step by step:

### Part (a): Solve \((x^2 + y^2)(p^2 + q^2) = 1\)

Given equation:

\[
(x^2 + y^2)(p^2 + q^2) = 1
\]

This equation suggests that the product of two sums of squares is equal to 1. There are different cases to consider for this product to equal 1:

#### Case 1:
Assume \( x^2 + y^2 = 1 \) and \( p^2 + q^2 = 1 \).

- In this case, both expressions represent the equations of circles in the \(xy\)-plane and \(pq\)-plane, respectively. The points \((x, y)\) and \((p, q)\) lie on the unit circles centered at the origin.

#### Case 2:
Assume \( x^2 + y^2 = 0 \) and \( p^2 + q^2 = \infty \).

- For this to be true, \(x\) and \(y\) must both be 0 (since the square of a real number is non-negative, and the only way their sum is zero is if each is zero), but \(p\) and \(q\) can be very large (approaching infinity).

#### Case 3:
Assume \( x^2 + y^2 = \infty \) and \( p^2 + q^2 = 0 \).

- Similarly, in this case, \(p\) and \(q\) must both be 0, while \(x\) and \(y\) can be very large.

These cases indicate that the general solutions are:

\[
(x, y, p, q) = \left( \cos \theta, \sin \theta, \cos \phi, \sin \phi \right)
\]

where \(\theta\) and \(\phi\) are any real numbers.

### Part (b): Solve the equation \( zpq = p + q \)

Given equation:

\[
zpq = p + q
\]

We can rearrange this to:

\[
zpq - p - q = 0
\]

Factor the equation:

\[
p(zq - 1) + q(zp - 1) = 0
\]

For this equation to hold true, the expression inside each parenthesis must be zero, or \(p\) and \(q\) must be such that their respective multipliers sum to zero.

### Case 1: \( p = 0 \) or \( q = 0 \)

- If \( p = 0 \) or \( q = 0 \), the equation reduces to \( 0 = 0 \), which is always true. However, this would imply that either \(p = 0\) or \(q = 0\) with the other factor non-zero, which leads to \( zp = 1 \) or \( zq = 1 \), respectively. So, \(z = \frac{1}{q}\) or \(z = \frac{1}{p}\).

### Case 2: \( zq - 1 = 0 \) and \( zp - 1 = 0 \)

- Solving these, we get \( z = \frac{1}{q} \) and \( z = \frac{1}{p} \). Therefore, for \(z\) to satisfy both equations simultaneously, \(p\) and \(q\) must be equal, i.e., \(p = q\).

Therefore, the solutions are:
- \(p = 0\), \(q\) is arbitrary, and \(z = \frac{1}{q}\) if \(q \neq 0\).
- \(q = 0\), \(p\) is arbitrary, and \(z = \frac{1}{p}\) if \(p \neq 0\).
- \(p = q\) and \(z = \frac{1}{p}\).

### Summary:
- For Part (a), solutions are \((x, y, p, q) = \left( \cos \theta, \sin \theta, \cos \phi, \sin \phi \right)\) for any real \(\theta\) and \(\phi\).
- For Part (b), solutions are \(p = 0\), \(z = \frac{1}{q}\); \(q = 0\), \(z = \frac{1}{p}\); or \(p = q\) and \(z = \frac{1}{p}\).

Would you like more details on any of the steps?

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Here are 5 related questions:
1. How do you interpret complex roots in equations like \((x^2 + y^2)(p^2 + q^2) = 1\)?
2. Can the general solution \((\cos \theta, \sin \theta, \cos \phi, \sin \phi)\) be applied to other types of equations?
3. What happens if the equation \(zp = q\) had additional terms or constraints?
4. Can the conditions from Part (a) be generalized to higher dimensions, like in the case of 3D spheres?
5. How do these solutions change if we consider only positive values for \(x, y, p,\) and \(q\)?

**Tip:** When factoring equations, always consider both straightforward and edge cases, as they can reveal different possible solutions.

Learn how to solve the equations (x^2 + y^2)(p^2 + q^2) = 1 and zpq = p + q with detailed step-by-step solutions. This problem covers algebraic equations involving squares, trigonometric identities, and circle equations. Suitable for students in Grades 10-12.

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