Sketch the domain of the pair of real numbers $(p,\ q)$ in the $pq$ plane such that the equation in $x:px^2+(p^2-q)x-(2p-q-1)=0$ has solution and the real part of all solutions are negative.

[i]1992 Tokyo University entrance exam/Humanities[/i]
To determine the domain of the pair of real numbers \((p, q)\) in the \(pq\) plane such that the equation in \(x\):
\[ px^2 + (p^2 - q)x - (2p - q - 1) = 0 \]
has solutions where the real part of all solutions are negative, we need to analyze the roots of the quadratic equation.

Let the roots of the quadratic equation be \(\alpha\) and \(\beta\). We will consider two cases: \(p = 0\) and \(p \neq 0\).

### Case 1: \(p = 0\)
When \(p = 0\), the equation simplifies to:
\[ (p^2 - q)x - (2p - q - 1) = 0 \]
\[ -qx - (q + 1) = 0 \]
\[ x = \frac{q + 1}{q} \]

For the real part of the solution \(x\) to be negative:
\[ \frac{q + 1}{q} < 0 \]

This inequality holds when:
\[ -1 < q < 0 \]

### Case 2: \(p \neq 0\)
For \(p \neq 0\), we need to ensure that the real parts of the roots \(\alpha\) and \(\beta\) are negative. We will use Vieta's formulas and the properties of quadratic equations.

#### Subcase (i): Discriminant \(D \geq 0\)
The discriminant \(D\) of the quadratic equation is given by:
\[ D = (p^2 - q)^2 + 4p(2p - q - 1) \]

For the roots to be real and the real parts to be negative, we need:
\[ \alpha + \beta = -\frac{p^2 - q}{p} < 0 \]
\[ \alpha \beta = -\frac{2p - q - 1}{p} > 0 \]

From \(\alpha + \beta < 0\):
\[ -\frac{p^2 - q}{p} < 0 \]
\[ p^2 - q > 0 \]
\[ q < p^2 \]

From \(\alpha \beta > 0\):
\[ -\frac{2p - q - 1}{p} > 0 \]
\[ 2p - q - 1 < 0 \]
\[ q > 2p - 1 \]

#### Subcase (ii): Discriminant \(D < 0\)
When the discriminant is negative, the roots are complex conjugates. For the real parts of the complex roots to be negative:
\[ 2 \text{Re}(\alpha) = 2 \text{Re}(\beta) = -\frac{p^2 - q}{p} < 0 \]
\[ \alpha \beta = -\frac{2p - q - 1}{p} > 0 \]

These conditions are the same as in subcase (i):
\[ q < p^2 \]
\[ q > 2p - 1 \]

### Conclusion
Combining the results from both cases, we get the domain of \((p, q)\) as:
\[ (p = 0 \text{ and } -1 < q < 0) \text{ or } (p > 0 \text{ and } q < p^2 \text{ and } q > 2p - 1) \]

The final answer is \( \boxed{ (p = 0 \text{ and } -1 < q < 0) \text{ or } (p > 0 \text{ and } q < p^2 \text{ and } q > 2p - 1) } \)