Question

The quadratic formula works whether the coefficients of the equation are real or complex. Solve these equations using the quadratic formula, and, if necessary, DeMoivre's Theorem. z²-i z+1=0

Answer 1

The quadratic equation z² - iz + 1 = 0 is solved using the standard quadratic formula. The roots are found by substituting a = 1, b = -i, and c = 1 into the formula. DeMoivre's Theorem isn't necessary as the discriminant is real in this instance.

Step-by-step explanation:

To solve the quadratic equation z² - iz + 1 = 0 using the quadratic formula, we apply the standard formula for quadratic equations of the form ax² + bx + c = 0, which is:

x = (-b ± √(b² - 4ac)) / (2a)

For the given equation, a = 1, b = -i, and c = 1. Substituting these values into the formula gives:

z = (i ± √((-i)² - 4(1)(1))) / (2(1))

z = (i ± √(-1 + 4)) / 2

z = (i ± √(3)) / 2

In this case, the discriminant (-i)² - 4(1)(1) results in a real number, so there's no need for DeMoivre's Theorem, which is required to find the roots of complex numbers. However, if in other scenarios the discriminant was complex, DeMoivre's Theorem can be very useful for calculating complex roots. The solutions to the equation are the two roots represented by the ± symbol in the quadratic formula.