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

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

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

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

[tex]z_1=\frac{2i+4}{2}=i+2[/tex] and [tex]z_1=\frac{2i-4}{2}=i-2[/tex]

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

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

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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