If a polynomial equation p(x)=0 has 3+4i as a solution
3-i
4-3i
4+3i
3-4i
As Damon said, complex roots come in conjugate pairs.
Using properties of roots:
sum of our roots = 3+4i + 3-4i = 6
product of roots = (3+4i)(3-4i)
= 9 - 16i^2 = 25
p(x) = x^2 - 6x + 25
if 3 + 4 i is a solution
then its complex conjugate
3 - 4 i
is also a solution
because when you solve a quadratic for example
(- b +/- sqrt(b^2-4ac))/2a
if b^2-4ac is negative
then you have
-b + number*i
and
-b - same number*i
They come in complex conjugate PAIRS
Are you looking for another solution?
if so what do you think/
To find the other complex solutions to the polynomial equation p(x) = 0, given that 3 + 4i is a solution, we can use the concept of complex conjugates.
1. Start with the given solution: 3 + 4i.
2. Take the complex conjugate of the given solution by changing the sign of the imaginary part: 3 - 4i. This is one of the other complex solutions.
3. Repeat the process for the complex conjugate solution: 3 - 4i.
- Complex conjugate of 3 - 4i: 3 + 4i. This is another complex solution.
4. Finally, we have two complex solutions: 3 - 4i and 3 + 4i.
Therefore, the correct complex solutions to the polynomial equation p(x) = 0 are:
- 3 + 4i
- 3 - 4i