Find the modulus and argument of y^2+4y+8=0

Answer 1

To find the modulus and argument of the complex number solutions to the equation y^2 + 4y + 8 = 0, we first need to solve for y.

Completing the square, we get:

y^2 + 4y + 4 = -4
(y + 2)^2 = -4

Taking the square root of both sides:

y + 2 = ±2i
y = -2 ± 2i

Now, we have two complex numbers as solutions:
y1 = -2 + 2i
y2 = -2 - 2i

To find the modulus of these complex numbers, we use the formula:
|z| = sqrt(Re(z)^2 + Im(z)^2)

For y1 = -2 + 2i:
|y1| = sqrt((-2)^2 + 2^2)
|y1| = sqrt(4 + 4)
|y1| = sqrt(8)
|y1| = 2√2

For y2 = -2 - 2i:
|y2| = sqrt((-2)^2 + (-2)^2)
|y2| = sqrt(4 + 4)
|y2| = sqrt(8)
|y2| = 2√2

Therefore, the modulus of both complex numbers is 2√2.

To find the argument of these complex numbers, we use the formula:
arg(z) = tan^(-1)(Im(z)/Re(z))

For y1 = -2 + 2i:
arg(y1) = tan^(-1)(2/-2)
arg(y1) = tan^(-1)(-1)
arg(y1) = -π/4

For y2 = -2 - 2i:
arg(y2) = tan^(-1)(-2/-2)
arg(y2) = tan^(-1)(1)
arg(y2) = π/4

Therefore, the argument of y1 is -π/4 and the argument of y2 is π/4.