maths complex numbers

Question

Show that z = 1 − 2i is a root of the equation z ^4 − 6z ^3 + 26z^2 − 46z + 65 = 0.

Answer 1

z = 1 -2i

is one zero

they come in conjugate pairs, so z=1+2i is another zero or root

change the sign of each zero and stick an z in front of it to get the factors

(x-(1+2i)(x+(1-2i) = x^2 -2z + 5

if z^2 -2z +5

divides evenly into the 4th degree polynomial, then the conjugate pair are zeros.

it divides evenly with a quotient z^2 -4z + 13

set it equal to zero to find the other two zeros

complete the square, the take the square root of both sides

z ^2 -4z + 4 = -13 +4

z-2 = + or -sqr(-9) = + or - 3i

= 2+3i and 2-3i

there are 4 imaginary zeros. the 4th degree polynomial never intersects the x axis

the graph is like a large W, with the minimum point above the x axis, and the ends of the W extending upwards to infinity.

using Descartes' method there are no negative zeros and either 0, 2 or 4 real positive zeros. Turns out there are no real zeros, positive or negative.