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- SOLUTION: I am having trouble with forming polynomials using real . . .
Question 464406: I am having trouble with forming polynomials using real coefficents: Degree: 4 Zeros: 4 multiplicity of 2, 2i I have to show the final fully multiplied polynomial Answer by Edwin McCravy (20048) (Show Source):
- SOLUTION: 1. Given: Z1= 2-2i , Z2= 3i and Z3= -3+i . Solve analytically . . .
Further Calculations** ** (c) Z1 * Z3** * Z1 * Z3 = (2 - 2i) * (-3 + i) * Z1 * Z3 = -6 + 2i + 6i - 2i² * Z1 * Z3 = -6 + 8i + 2 (since i² = -1) * Z1 * Z3 = -4 + 8i ** (d) Z3 x Z2** * The cross product is not defined for complex numbers in the same way it is for vectors in 3D space ** (e) Acute Angle between Z1 and Z2** * Find the magnitudes
- SOLUTION: Find all complex solutions to the equation z^3 = 8i - (15 . . .
You can put this solution on YOUR website! Here's how to find the complex solutions to the equation z³ = 8i - (15 + 2i)z² + (7 - 4i)z: 1 **Rewrite the equation:** Move all terms to one side to get a standard polynomial form: z³ + (15 + 2i)z² - (7 - 4i)z - 8i = 0 2 **Solve the cubic equation:** Cubic equations can be challenging to solve by hand There isn't a simple, general formula like
- SOLUTION: Does 2i times the square root of -4 equal -4? I think 2i . . .
I think 2i times the square root of -4 does equal -4, because the square root of -4 equals 2i and (2i) (2i) equals 4i^2 i^2 = -1 -> -> SOLUTION: Does 2i times the square root of -4 equal -4? I think 2i times the square root of -4 does equal -4, because the square root of -4 equals 2i and (2i) (2i) equals 4i^2 i^2 = -1
- SOLUTION: Solve the equation z^2 + (2i - 3)z + 5 - i = 0.
You can put this solution on YOUR website! In the quadratic equation ax^2+bx+c = 0 the sum of the roots is -b a and the product is c a This holds for quadratic equations with complex coefficients With a=1 in this equation, the sum of the roots is - (2i-3) = 3-2i and the product is 5-i Let the roots be p+qi and r+si Then their product is (pr-qs)+ (ps+qr)i and their sum is (p+r)+ (q+s)i So
- SOLUTION: A sequence of positive integers with a_1 = 1 and a_9 + a_ {10 . . .
Question 1176668: A sequence of positive integers with a_1 = 1 and a_9 + a_ {10} = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all i>=1, the terms a_ {2i - 1}, a_ {2i}, a_ {2i + 1} are in geometric progression, and the terms a_ {2i}, a_ {2i + 1}, and a_ {2i + 2} are in
- How do you plot -3 + 2i? | Socratic
How do you plot −3 + 2i? Precalculus Complex Numbers in Trigonometric Form Complex Number Plane
- SOLUTION: form a polynomial f (x) with real coefficients having the . . .
You can put this solution on YOUR website! A polynomial of real coefficients will have as many zeros as the degree of the polynomial Also, any complex zeros will come in conjugate pairs So if 1-2i is a zero, then 1+2i will also be a zero So your 4th degree polynomial will have zeros of -1, 2, 1-2i and 1+2i If z is a zero of a polynomial, then (x-z) is a factor of the polynomial So f (x
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