▸ ▸ » Complex numbers equation

Complex numbers equation

Asked Nov 13, 2010, 08:20 AM — 1 Answer

How to solve $x^4-(5-2i)x^2-10i=0$ with complex numbers?

Search this Thread

Share |

1 Answer

galactus Posts: 2,272, Reputation: 1436

Ultra Member

Nov 13, 2010, 09:04 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:

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