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EleoNora [17]
3 years ago
10

The quadratic formula works whether the coefficients of the equation are real or complex. Solve the following equations using th

e quadratic formula and, if necessary, De Moivre's Theorem. (Enter your answers as a comma-separated list.)
z2 − 2iz − 5 = 0
Mathematics
1 answer:
RoseWind [281]3 years ago
6 0

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.  

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