Hard maths question about roots of equations?

Question

Verify that 1 + i is a root of the equation (z^4) - 6(z^3) + 15(z^2) - 18z + 10 = 0
And hence find all the roots.

Cheers!!

Answer 1

You may do the verification in one of two ways. The first is to actually substitute 1+i for z, and see that it matches zero. The other method is to use synthetic division and the factor theorem to test it -- I like this method better, simply because we get a factorization out of the method for free if 1+i is a root. So doing this:

1+i| 1 -6 ...15 .. -18 ..... 10
......... 1+i -6-4i 13+5i.. -10
-------------------------------------
.......1 -5+i 9-4i -5+5i ...... 0

So indeed, 1+i is a factor, and the remaining polynomial is z³ + (-5+i)z² + (9-4i)z + (-5+5i). Now, the original polynomial had only real coefficients, so if a complex number is a root of the polynomial, then so too is its conjugate. So we know 1-i is going to be a factor, so let's divide it out:

1-i| 1 -5+i 9-4i -5+5i
......... 1-i -4+4i 5-5i
-----------------------------
...... 1 -4 .... 5 .... 0

So the remaining polynomial is z² - 4z + 5 = 0. Then all that remains is to solve for the roots of this polynomial, and you will then have all the roots of the original quartic. So:

z = (4±√(16-20))/2 = 2±i

Therefore, the roots of the original quartic are 1+i, 1-i, 2+i, and 2-i.

Answer 2

The word conjugate roots springs to mind from memory 30 odd years ago. If there is a complex root Z (a+bi) of a polynomial (not to be confused with the variable z above) then the conjugate of Z (a-bi) also has to be a root. Multiply these two roots together and divide the original polynomial by these to be left with a quadratic in z with real coefficients. Then find the roots of this if necessary employing the formula for solving a quadratic.

Answer 3

Here is what you do:
Let z = 1 + i, then plug into the LHS and see if it indeed does equal zero. This exercise is best done by you to get used to working with complex powers. Be careful of the signs.