4) The roots of the equation z^2+2z+4=0 are denoted by α and β.
a) find α and β in the form re^iθ, giving the values of r and θ
b) Using de Moivre's Theorem, show that α^3=β^3.
c) find the exact value of αβ*+α*β
z = (-2±√-12)/2 = -1±i√3
or 6e^(i 2π/3) and 6e^(i 4π/3)
now the rest is easy...right?
a) To find α and β in the form re^iθ, we can use the quadratic formula. The quadratic formula states that for an equation of the form az^2 + bz + c = 0, the solutions for z are given by:
z = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = 2, and c = 4. Plugging these values into the quadratic formula, we have:
z = (-2 ± √(2^2 - 4(1)(4))) / (2(1))
z = (-2 ± √(4 - 16)) / 2
z = (-2 ± √(-12)) / 2
Since we have a negative under the square root, we can write it as:
z = (-2 ± i√12) / 2
z = -1 ± i√3
Therefore, α = -1 + i√3 and β = -1 - i√3.
To convert these complex numbers to the polar form re^iθ, we can find the magnitude r and the argument θ:
r = √((-1)^2 + (√3)^2) = √(1 + 3) = 2
θ = tan^(-1)(√3 / -1) = tan^(-1)(-√3) = -π / 3
So α = 2e^(i(-π / 3)) and β = 2e^(i(-π / 3)).
b) Let's calculate α^3 and β^3 using de Moivre's Theorem. According to de Moivre's Theorem, for any complex number z and any positive integer n, we have:
z^n = r^n e^(inθ)
Therefore:
α^3 = (2e^(i(-π / 3)))^3
= 2^3 (e^(i(-π / 3))^3
= 8e^(i(-π))
Similarly:
β^3 = (2e^(i(-π / 3)))^3
= 2^3 (e^(i(-π / 3))^3
= 8e^(i(-π))
Since α^3 = β^3, we have shown that α^3 = β^3.
c) To find the value of αβ* + α*β, we need to calculate the complex conjugates of α and β, denoted as α* and β* respectively.
α* = 2e^(-i(-π / 3))
= 2e^(iπ / 3)
β* = 2e^(-i(-π / 3))
= 2e^(iπ / 3)
Now, we can calculate αβ* + α*β:
αβ* + α*β = (2e^(i(-π / 3)))(2e^(iπ / 3)) + (2e^(iπ / 3))(2e^(i(-π / 3)))
= 4 e^(i(-π / 3) + iπ / 3) + 4 e^(iπ / 3 + i(-π / 3))
= 4 e^(0) + 4 e^(0)
= 4 + 4
= 8
Therefore, the exact value of αβ* + α*β is 8.
a) To find the roots α and β in the form re^iθ, we can use the quadratic formula. The quadratic formula states that for an equation of the form az^2 + bz + c = 0, the roots can be found using the formula:
z = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the equation is z^2 + 2z + 4 = 0, so a = 1, b = 2, and c = 4. Plugging these values into the quadratic formula, we get:
z = (-2 ± √(2^2 - 4(1)(4))) / (2(1))
= (-2 ± √(-12)) / 2
= (-2 ± √(12)i) / 2
Simplifying further, we have:
z = (-1 ± √3i)
The roots α and β are represented by these values. Let's assign α = (-1 + √3i) and β = (-1 - √3i).
To express these in the form re^iθ, we need to find the values of r and θ.
For α:
r = √((-1)^2 + (√3)^2) = √(1 + 3) = 2
θ = arctan(√3 / -1) = -π/3 (taking the angle in the third quadrant)
So, α is represented as 2e^(-iπ/3).
For β:
r = √((-1)^2 + (-√3)^2) = √(1 + 3) = 2
θ = arctan(-√3 / -1) = π/3 (taking the angle in the first quadrant)
So, β is represented as 2e^(iπ/3).
b) To show that α^3 = β^3 using de Moivre's Theorem, we need to cube both α and β and show that they are equal.
Applying de Moivre's Theorem to α:
α^3 = (2e^(-iπ/3))^3
= 2^3 e^(-iπ)
= 8e^(-iπ)
= 8(cos(-π) + i sin(-π))
= 8(-1 + 0i)
= -8
Applying de Moivre's Theorem to β:
β^3 = (2e^(iπ/3))^3
= 2^3 e^(3iπ)
= 8e^(3iπ)
= 8(cos(3π) + i sin(3π))
= 8(-1 + 0i)
= -8
Therefore, α^3 = β^3 = -8.
c) To find the exact value of αβ* + α*β, we need to multiply α by the complex conjugate of β and add it to the product of the conjugate of α and β.
αβ* = (2e^(-iπ/3)) * (2e^(iπ/3))
= 4e^0
= 4
α*β = (2e^(-iπ/3)) * (2e^(iπ/3))
= 4e^0
= 4
Therefore, αβ* + α*β = 4 + 4 = 8.