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lemon14
Nov 13, 2010, 09:20 AM
How to solve x^4-(5-2i)x^2-10i=0 with complex numbers?

galactus
Nov 13, 2010, 10:04 AM
We can factor just like we would a regular old quadratic.

x^{4}-(5-2i)x^{2}-10i=0

What two numbers when multiplied equal -10i and when added equal -5+2i?

How about 2i and -5?

x^{4}+2ix^{2}-5x^{2}-10i

Group and factor

x^{2}(x^{2}+2i)-5(x^{2}+2i)

(x^{2}-5)(x^{2}+2i)=0

What values of x make

x^{2}-5=0?

Difference of two squares:

(x+\sqrt{5})(x-\sqrt{5})=0

What values of x make

x^{2}+2i=0?

Factor:

(x-(1-i))(x+(1-i))=0