answer:To solve the quadratic equation 2z^2 - (7+4i)z + (6-9i) = 0, we can use the quadratic formula: z = (-b ± √(b^2 - 4ac)) / 2a In this case, a = 2, b = -(7+4i), and c = (6-9i). First, let's find the discriminant, Δ = b^2 - 4ac: Δ = (-(7+4i))^2 - 4(2)(6-9i) Δ = (7+4i)^2 - 8(6-9i) Δ = (49 + 56i + 16i^2) - (48 - 72i) Δ = (49 + 56i - 16) - (48 - 72i) Δ = (33 + 128i) Now, we can find the two solutions for z using the quadratic formula: z1 = (b + √Δ) / 2a z1 = ((7+4i) + √(33 + 128i)) / 4 z2 = (b - √Δ) / 2a z2 = ((7+4i) - √(33 + 128i)) / 4 Unfortunately, the square root of the discriminant, √(33 + 128i), cannot be simplified further without using numerical approximations. Therefore, the solutions for z are: z1 = ((7+4i) + √(33 + 128i)) / 4 z2 = ((7+4i) - √(33 + 128i)) / 4

answer:To solve the quadratic equation z^2 + 4z + 5 = 0, we can use the quadratic formula: z = (-b ± √(b^2 - 4ac)) / 2a In this case, a = 1, b = 4, and c = 5. Plugging these values into the formula, we get: z = (-4 ± √(4^2 - 4(1)(5))) / 2(1) z = (-4 ± √(16 - 20)) / 2 z = (-4 ± √(-4)) / 2 Since the discriminant (the value inside the square root) is negative, we will have complex solutions. We can rewrite the square root as: √(-4) = √(4 * -1) = 2√(-1) = 2i Now, we can find the two complex solutions: z = (-4 + 2i) / 2 = -2 + i z = (-4 - 2i) / 2 = -2 - i So, the solutions to the equation z^2 + 4z + 5 = 0 are z = -2 + i and z = -2 - i.

answer:To solve the quadratic equation z^2 + 4z + 13 = 0, we can use the quadratic formula: z = (-b ± √(b^2 - 4ac)) / 2a In this case, a = 1, b = 4, and c = 13. Plugging these values into the formula, we get: z = (-4 ± √(4^2 - 4(1)(13))) / 2(1) z = (-4 ± √(16 - 52)) / 2 Since the discriminant (b^2 - 4ac) is negative, we know that the solutions will be complex numbers. We can rewrite the square root of a negative number as the square root of the positive number times the imaginary unit i: z = (-4 ± √(-36)) / 2 z = (-4 ± √(36) * i) / 2 z = (-4 ± 6i) / 2 Now, we can simplify the expression: z = -2 ± 3i So, the complex number solutions for the equation z^2 + 4z + 13 = 0 are: z = -2 + 3i and z = -2 - 3i

answer:To solve the complex number equation, we can use the quadratic formula: z = (-b ± √(b^2 - 4ac)) / 2a In this case, a = 1, b = (3 - i), and c = -(4 + 2i). Now, we can plug these values into the quadratic formula: z = (-(3 - i) ± √((3 - i)^2 - 4(1)(-(4 + 2i)))) / 2(1) First, let's simplify the discriminant (the part inside the square root): (3 - i)^2 - 4(1)(-(4 + 2i)) = (9 - 6i + i^2) - (-8 - 4i) = (9 - 6i - 1) + 8 + 4i = 16 - 2i Now, we can plug the discriminant back into the quadratic formula: z = (-(3 - i) ± √(16 - 2i)) / 2 Next, we need to find the square root of the discriminant. To do this, we can rewrite the complex number in polar form and use De Moivre's theorem. The modulus of the discriminant is: |16 - 2i| = √(16^2 + (-2)^2) = √(256 + 4) = √260 The argument of the discriminant is: arg(16 - 2i) = arctan(-2/16) = -arctan(1/8) Now, we can rewrite the discriminant in polar form: 16 - 2i = √260 * (cos(-arctan(1/8)) + i * sin(-arctan(1/8))) Using De Moivre's theorem, we can find the square root of the discriminant: ±√(16 - 2i) = ±√(√260 * (cos(-arctan(1/8)) + i * sin(-arctan(1/8)))) = ±(√(√260) * (cos(-arctan(1/8)/2) + i * sin(-arctan(1/8)/2))) Now, we can plug the square root of the discriminant back into the quadratic formula: z = (-(3 - i) ± (√(√260) * (cos(-arctan(1/8)/2) + i * sin(-arctan(1/8)/2)))) / 2 Finally, we have the two solutions for z: z1 = (-(3 - i) + (√(√260) * (cos(-arctan(1/8)/2) + i * sin(-arctan(1/8)/2)))) / 2 z2 = (-(3 - i) - (√(√260) * (cos(-arctan(1/8)/2) + i * sin(-arctan(1/8)/2)))) / 2