Complex numbers α and β satisfy
αα¯=ββ¯=−17,α+β=7i.
What is the value of αβ?
α¯=conjugate of α
β¯=conjugate of β
To find the value of αβ, we can use the given information that αα¯ = ββ¯ = -17 and α + β = 7i.
Let's start by expressing α and β in terms of their real and imaginary parts. We can write α as α = a + bi and β as β = c + di, where a, b, c, and d are real numbers.
Since αα¯ = -17, we have (a + bi)(a - bi) = -17.
Expanding the expression, we get a^2 + b^2 = -17. (Equation 1)
Similarly, ββ¯ = -17 gives us (c + di)(c - di) = -17.
Expanding this expression, we get c^2 + d^2 = -17. (Equation 2)
Now, we know that α + β = 7i. Substituting the values of α and β, we have (a + bi) + (c + di) = 7i.
Equating the real and imaginary parts on both sides of the equation, we get:
(a + c) + (b + d)i = 0 + 7i.
This gives us two equations:
a + c = 0 (Equation 3)
b + d = 7 (Equation 4)
Now, we need to solve Equations 1, 2, 3, and 4 simultaneously to find the values of a, b, c, and d.
From Equation 3, we have a = -c. Substituting this into Equation 1, we get (-c)^2 + b^2 = -17.
Simplifying the equation, we have c^2 + b^2 = -17. (Equation 5)
Similarly, from Equation 4, we have b = 7 - d. Substituting this into Equation 5, we get c^2 + (7 - d)^2 = -17.
Expanding and simplifying the equation, we have c^2 + d^2 - 14d + 50 = -17.
Rearranging the equation, we get c^2 + d^2 - 14d + 67 = 0. (Equation 6)
Now we have a system of two equations: Equation 2 and Equation 6.
c^2 + d^2 = -17 (Equation 2)
c^2 + d^2 - 14d + 67 = 0 (Equation 6)
Subtracting Equation 2 from Equation 6, we eliminate the term c^2 + d^2:
(c^2 + d^2 - 14d + 67) - (c^2 + d^2) = 0 - (-17).
Simplifying the equation, we get -14d + 67 = 17.
Rearranging the equation, we have -14d = 17 - 67.
Simplifying further, we get -14d = -50.
Dividing both sides by -14, we find d = -50/-14 = 25/7.
Substituting the value of d into Equation 4, we get b + 25/7 = 7.
Rearranging the equation, we have b = 7 - 25/7 = 49/7 - 25/7 = 24/7.
We have found the values of b and d. Now, substituting these values into Equation 3, we find a = -c.
Using Equation 3 with -c = a, we have -c + c = 0.
Thus, a = 0.
Finally, we can express α and β in terms of their real and imaginary parts:
α = a + bi = 0 + (24/7)i = (24/7)i.
β = c + di = -c + (25/7)i = -(25/7)i.
Therefore, αβ = (24/7)i * -(25/7)i = -(24/7 * 25/7)i^2 = -(600/49)(-1) = 600/49.
So, the value of αβ is 600/49.