We recall the property of complex roots: If a complex number is the root of an equation, then it's conjugate is also the root of the equation.
Fot the beginning, you'v get as roots 2 complex numbers: z and z'(z' is the conjugate of z).
z = a + bi
z' = a - bi
Now,l we'll verify if the complex numbers are the roots of the equation by substituting them into the original equation.
(1+3i)^3 - (1+3i)^2 + 8(1+3i) + 10 = 0
We'll expand the cube using the formula:
(a+b)^3 = a^3 + b^3 + 3ab(a+b)
a = 1 and b = 3i
(1+3i)^3 = 1^3 + (3i)^3 + 3*1*3i*(1+3i)
(1+3i)^3 = 1 - 27i + 9i(1+3i)
We'll remove the brackets:
(1+3i)^3 = 1 - 27i + 9i - 27
We'll combine real parts and imaginary parts:
(1+3i)^3 = -26 + i(9-27)
(1+3i)^3 = -26 - 18i
We'll expand the square using the formula:
(a+b)^2 = a^2 + 2ab + b^2
(1+3i)^2 = 1^2 + 2*1*3i + (3i)^2
(1+3i)^2 = 1 + 6i - 9
(1+3i)^2 = -8 + 6i
We'll substitute the results into the expression (1+3i)^3 - (1+3i)^2 + 8(1+3i) + 10 = 0.
-26 - 18i - (-8 + 6i) + 8 + 24i + 10 = 0
We'll combine like terms:
(-26 + 8 + 8 + 10) + i(-18 - 6 + 24) = 0
0 + 0*i = 0
It is obvious that 1 + 3i is the root of the equation.
According to the rule, the conjugate of 1 + 3i, namely 1 - 3i, is also the root of the equation. So it is not necessary to verify if 1 - 3i is the root, since we've demonstrated that 1 + 3i is the root of the equation.
Conclusion: 1 + 3i and 1 - 3i are the roots of the equation
Note: In calculus, we've substituted i^2 by it's value, -1.
To check whether 3i+1and 1-3i are the the solution of the equation,
of x^3-x^2+8x+10 = 0.
We know that if a+bi is a solution, then the conjugate a-bi is also a solution and there has to be a real root for a cubic. So let c be the real root.
We presume that 1+3i and 1-3i are the roots. Then the sum of the roots = - (-1) = 1.
Therefore 1+3i+1-3i+c =1
So c = -1.
So if 1+3i 1nd 1-3i are solutions, then their product is (1+3i)(1-3i) = 1-9(i^2) = 1+9 = 10. So the other root is = -1. As product of all 3 roots (1+3i)(1-3i)c = 10c = -10/1 as per the relation between the roots and coefficients. So c= -1.
We check now whether -1 is a root.
So f(-1) = (-1)^3-(1)^2 +8(-1)+10 = -1-1-8+10 = 0.
Also from the relation between the roots and coefficients, the sum of the roots = - coefficient of x^2/coefficient of x^3 = -(-1) =1.
That proves that 1+3i and 1-3i are the roots of x^3-x^2+8x+10 = 0