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- How do you divide (5i) (6+8i)? | Socratic
The answer is =0 4+0 3i The division of complex numbers is z_1 z_2 You multiply the numerator and denominator by the conjugate of the denominator (z_1*barz_2) (z_2*barz_2) If z=a+ib Then, barz=a-ib and i^2=-1 (a+b) (a-b)=a^2-b^2 Here, (5i) (6+8i)= ( (5i) (6-8i)) ( (6+8i) (6-8i)) = (30i-40i^2) (36-64i^2) = (40+30i) 100 =0 4+0 3i
- Multiplication of Complex Numbers - Socratic
Questions and Videos on Multiplication of Complex Numbers, within Precalculus
- How do you solve 11x ^ { 2} + 10= 2x ^ { 2} - 15? | Socratic
x = +-5 3i Given: 11x^2+10 = 2x^2-15 Subtract the right hand side from the left to get: 9x^2+25 = 0 Now x^2 >= 0 for all real values of x, so this has no real solutions
- What is all of the real and imaginary zeros of #y= (x^2-9 . . . - Socratic
We have 4 zeros, 3 with multiplicity 3 and -3,3i and -3i with multiplicity of 1 y= (x^2-9) (x^2+9) (x-3)^2 = (x^2-3^2) (x^2- (3i)^2) (x-3)^2 = (x+3) (x-3) (x-3i) (x+
- How do you write x^4 + 13x^2 + 36 as a product of linear factors . . .
Therefor; Y = (X + 4) (X + 9) y = (x^2 + 4) (x^2 + 9) To transform y to linear factors, we can use complex numbers, with i^2 = -1 y = (x - 2i) (x + 2i) (x - 3i) (x + 3i)
- Site Map - Vector Operations Questions and Videos | Socratic
Given the displacement vectors #A= (3i – 4j + 4k)m#, and #B= (2i + 3j – 7k)m#, how do you find the magnitudes of the vectors #C= A+B#? Vector A has a magnitude of 13 units at a direction of 250 degrees and vector B has a magnitude of 27 units at 330 degrees, both measured with respect to the positive x axis What is the sum of A and B?
- Question #d629b - Socratic
b) I multiply the 2 brackets: #4*7-4*3i+7*3i-3*3i^2=28-12i+21i+9=37+9i# Answer link
- Powers of Complex Numbers Questions and Videos - Socratic
Questions and Videos on Powers of Complex Numbers, within Precalculus
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