For a complex number z, the product of the real parts of the roots of the equation `z^(2)-z=5-5i` is (where, `i=sqrt(-1)`)

Options

A. `-25`
B. `-6`
C. `-5`
D. 25

Answer

To solve the problem, we need to find the roots of the equation \( z^2 - z = 5 - 5i \) and then calculate the product of the real parts of these roots.

### Step 1: Rearranging the equation
We start with the equation:
\[ z^2 - z - (5 - 5i) = 0 \]
This can be rewritten as:
\[ z^2 - z - 5 + 5i = 0 \]

### Step 2: Using the quadratic formula
The standard form of a quadratic equation is \( az^2 + bz + c = 0 \). Here, \( a = 1 \), \( b = -1 \), and \( c = -5 + 5i \). We can use the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-5 + 5i)}}{2 \cdot 1} \]
This simplifies to:
\[ z = \frac{1 \pm \sqrt{1 + 20 - 20i}}{2} \]
\[ z = \frac{1 \pm \sqrt{21 - 20i}}{2} \]

### Step 3: Simplifying the square root
Next, we need to simplify \( \sqrt{21 - 20i} \). We can express it in the form \( a + bi \) where:
\[ (a + bi)^2 = 21 - 20i \]
Expanding this gives:
\[ a^2 - b^2 + 2abi = 21 - 20i \]
From this, we can set up the equations:
1. \( a^2 - b^2 = 21 \)
2. \( 2ab = -20 \)

From the second equation, we can express \( b \) in terms of \( a \):
\[ b = \frac{-20}{2a} = \frac{-10}{a} \]

### Step 4: Substituting \( b \) into the first equation
Substituting \( b \) into the first equation:
\[ a^2 - \left(\frac{-10}{a}\right)^2 = 21 \]
This simplifies to:
\[ a^2 - \frac{100}{a^2} = 21 \]
Multiplying through by \( a^2 \) to eliminate the fraction:
\[ a^4 - 21a^2 - 100 = 0 \]
Let \( x = a^2 \). The equation becomes:
\[ x^2 - 21x - 100 = 0 \]

### Step 5: Solving the quadratic for \( x \)
Using the quadratic formula:
\[ x = \frac{21 \pm \sqrt{21^2 + 4 \cdot 100}}{2} \]
Calculating the discriminant:
\[ 21^2 + 400 = 441 + 400 = 841 \]
Thus:
\[ x = \frac{21 \pm 29}{2} \]
Calculating the two possible values:
1. \( x = \frac{50}{2} = 25 \)
2. \( x = \frac{-8}{2} = -4 \) (not valid since \( x = a^2 \))

So, \( a^2 = 25 \) implies \( a = 5 \) or \( a = -5 \). We take \( a = 5 \).

### Step 6: Finding \( b \)
Substituting \( a = 5 \) back to find \( b \):
\[ 2(5)b = -20 \implies b = -2 \]
Thus, \( \sqrt{21 - 20i} = 5 - 2i \).

### Step 7: Finding the roots
Now substituting back into our expression for \( z \):
\[ z = \frac{1 \pm (5 - 2i)}{2} \]
Calculating the two roots:
1. \( z_1 = \frac{1 + 5 - 2i}{2} = \frac{6 - 2i}{2} = 3 - i \)
2. \( z_2 = \frac{1 - (5 - 2i)}{2} = \frac{1 - 5 + 2i}{2} = \frac{-4 + 2i}{2} = -2 + i \)

### Step 8: Finding the product of the real parts
The real parts of the roots are \( 3 \) and \( -2 \). The product is:
\[ 3 \times (-2) = -6 \]

### Final Answer
The product of the real parts of the roots is \( \boxed{-6} \).