For a complex number z, the equation `z^(2)+(p+iq)z r+" is "=0` has a real root (where p, q, r, s are non - zero real numbers and `i^(2) = -1`), then

Options

A. `pqr=r^(2)+p^(2)s`
B. `prs=q^(2)+r^(2)p`
C. `prs=p^(2)+s^(2)q`
D. `pqs=s^(2)+q^(2)r`

Answer

To solve the problem, we need to analyze the given equation for a complex number \( z \):

\[ z^2 + (p + iq)z + (r + is) = 0 \]

where \( p, q, r, s \) are non-zero real numbers, and \( i^2 = -1 \). We are tasked with determining the condition under which this equation has a real root.

### Step 1: Assume \( z \) is real
Since we are looking for a real root, we can let \( z = x \), where \( x \) is a real number.

### Step 2: Substitute \( z \) into the equation
Substituting \( z = x \) into the equation gives:

\[ x^2 + (p + iq)x + (r + is) = 0 \]

### Step 3: Separate real and imaginary parts
This expands to:

\[ x^2 + px + r + i(qx + s) = 0 \]

For this equation to hold, both the real part and the imaginary part must separately equal zero. Thus, we can write:

1. Real part:
\[ x^2 + px + r = 0 \]

2. Imaginary part:
\[ qx + s = 0 \]

### Step 4: Solve the imaginary part for \( x \)
From the imaginary part, we can solve for \( x \):

\[ qx + s = 0 \implies x = -\frac{s}{q} \]

### Step 5: Substitute \( x \) back into the real part
Now we substitute \( x = -\frac{s}{q} \) into the real part equation:

\[ \left(-\frac{s}{q}\right)^2 + p\left(-\frac{s}{q}\right) + r = 0 \]

This simplifies to:

\[ \frac{s^2}{q^2} - \frac{ps}{q} + r = 0 \]

### Step 6: Multiply through by \( q^2 \) to eliminate the denominator
Multiplying the entire equation by \( q^2 \) gives:

\[ s^2 - psq + rq^2 = 0 \]

### Step 7: Rearranging the equation
Rearranging this, we have:

\[ s^2 - psq + rq^2 = 0 \]

This can be rewritten as:

\[ pqs = s^2 + rq^2 \]

### Conclusion
Thus, we find that for the equation to have a real root, the condition must be:

\[ pqs = s^2 + rq^2 \]

### Final Answer
The correct option is \( \text{D: } pqs = s^2 + rq^2 \).