Find the magnitude of the following Vector: v= -2i - 4j ?
The magnitude of a vector v is given by the following formula: For v = (v1, v2) ||v|| = sqrt( (v1)^2 + (v2)^2) ) Therefore, since v = (-2, -4), ||v|| = sqrt ( (-2)^2 + (-4)^2 ) = sqrt (4 + 16) = sqrt(20), which can be reduced to 2sqrt(5)
How to find the other zero of this problem?
Since 1 - 2i is a zero of the function and since each term has real coefficients, then the conjugate of 1 - 2i must be a zero as well (x - (1 - 2i)) * (x - (1 + 2i)) * k = x^4 - 2x^3 + 6x^2 - 2x + 5 (x^2 - x * (1 + 2i) - x * (1 - 2i) + (1 - 2i) * (1 + 2i)) * k = x^4 - 2x^3 + 6x^2 - 2x + 5 (x^2 - x - 2ix - x + 2ix + 1 - 4i^2) * k =>> (x^2 - 2x + 1 + 4) * k => (x^2 - 2x + 5) * k Divide x^4 - 2x^3 + 6x^2 - 2x + 5 by x^2 - 2x + 5 (or factor by grouping): x^4 - 2x^3 + 5x^2 + x^2 - 2x + 5 => x^2 * (x^2 - 2x + 5) + 1 * (x^2 - 2x + 5) => (x^2 + 1) * (x^2 - 2x + 5) x^2 + 1 = 0 x^2 = -1 x = +/- sqrt(-1) x = -i , i x^2 - 2x + 5 = 0 x = (2 +/- sqrt(4 - 20)) / 2 x = (2 +/- sqrt(-16)) / 2 x = (2 +/- 4i) / 2 x = 1 +/- 2i x = -i , i , 1 - 2i , 1 + 2i
Find the Quotient of the complex number: 6/7+2i?
6/(7 + 2i) ::: multiply with conjugate of the denominator [6*(7-2i)] / [(7 + 2i)*(7 - 2i)] ... = [42 - 12i] / [49 + 4] ... = 42/53 - 12/53 i and for the second problem --- Quickmath was right
Find the product of the complex number: (5-2i)(5+2i)?
The product of two conjugate complex numbers is real: (5 - 2i) (5 + 2i) = 5 * 5 + 5 * 2i - 5 * 2i - 2i * 2i ... = 25 + 10i - 10i - 4i^2 ... = 25 + 4 = 29 ------------- sqrt(2y^2 - 1) = y -------- take squares (introduces extraneous sol.) 2y^2 - 1 = y -------- rearrange 2y^2 - y - 1 = 0 -------- solve using quadratic formula y = 1/4 +/- 1/4 sqrt 9 y = 1 or y = -1/2 Check: sqrt (2 * 1^2 - 1) = sqrt 1 = 1, so y = 1 is solution sqrt (2*(-1/2)^2 - 1) = sqrt (-1/2) <> -1/2 --> extraneous solution