Question

The quadratic formula works whether the coefficients of the equation are real or complex. Solve the following equations using the quadratic formula and, if necessary, De Moivre's Theorem. (Enter your answers as a comma-separated list.)
z2 − 2iz − 5 = 0

Answer 1

Answer:

z=i+2

,

z=i-2

Step-by-step explanation:

Remember that the quadratic equation $az^2+bz+c=0$ has the complex solutions $z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (this is the quadratic formula).

Apply this with a=1, b=-2i and c=-5 to get:

$z=\frac{2i\pm \sqrt{(-2i)^2-4(-5)}}{2}=\frac{2i\pm \sqrt{-4+20}}{2}=\frac{2i\pm 4}{2}$. Then the solutions are

$z_1=\frac{2i+4}{2}=i+2$ and $z_1=\frac{2i-4}{2}=i-2$

It isn't necessary to use De Moivre's formula.