Suppose a quadratic function f(z) where z is complex has two complex roots, then f < 0, as the discriminant $$b^2-4ac < 0$$ implies f(z) has two complex roots. Therefore you can apply real operators to complex numbers
Disprove/explain why it is wrong.
Prove my pondering wrong :)?
2017-03-28 6:10 am
回答 (3)
2017-03-28 6:41 am
✔ 最佳答案
Complex numbers are of the form: a + bi where a and b are real numbers and i is the imaginary number (=√(-1))
Note that the discriminant really applies only to polynomials such as (ax^2 + bx + c) where a, b and c are real numbers.
2017-03-28 6:28 am
But you are NOT applying operators to complex numbers. You are applying them to the coefficients of the quadratic equation, which are real numbers.
2017-03-28 6:16 am
counter example
f(z) =z^2+1 > 0
and f(z)=0 --->z = -i or i
f(z) =z^2+1 > 0
and f(z)=0 --->z = -i or i
收錄日期: 2021-04-24 00:22:09
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