How to find the solution of a quadratic equation with complex coefficients?
I know how to find the solution for a quadratic equation with real coefficients. But if the coefficient changes to complex numbers then what is the change in the solution? Want an example of such equation with solution.
$\endgroup$ 42 Answers
$\begingroup$It's no different. The quadratic formula works regardless of whether the coefficients are real or complex.
Consider the example $$(3+i)x^2 + (2-i)x + (5+2i) = 0$$
The quadratic formula gives
$$x = \frac{-(2-i) \pm \sqrt{ (2-i)^2-4(3+i)(5+2i) } }{2(3+i)}$$
Simplifying this is kind of a pain, of course. Under the radical you have to multiply everything out and combine terms. Eventually you get the radical into the form $\sqrt{M+Ni}$ where $M$ and $N$ are some constants -- in this example, they will be integers. Then the question is, how do you simplify the square root of a complex number? (For that, see How do I get the square root of a complex number?).
$\endgroup$ $\begingroup$Let's see if this works. Kind of unclear how to use these tools, so I just was going to snapshot a quick writeup.
