Prove that 1+i is a root of the equation z^4+3z^2-6z+10=0. Find all the other roots.
$\displaystyle z^4+3z^2-6z+10=0$Originally Posted by kingkaisai2
$\displaystyle (1+i)^4+3(1+i)^2-6(1+i)+10$
$\displaystyle (1^4+4*1^3*i+6*1^2*i^2+4*1*i^3+i^4)$$\displaystyle +3(1^2+2*1*i+i^2)-6(1+i)+10$
$\displaystyle (1 + 4i - 6 - 4i + 1)+ 3(1 + 2i - 1) - 6(1 + i) + 10$
$\displaystyle -4 + 6i - 6 - 6i + 10$
$\displaystyle 0$ (Check.)
Now, if 1+i is a root, so is 1-i. Thus we know that $\displaystyle \left ( z - (1+i) \right ) \left ( z - (1-i) \right )$$\displaystyle = \left ( (z-1) - i \right ) \left ( (z-1) + i \right )$$\displaystyle = (z-1)^2 - i^2 = z^2 - 2z + 1 + 1 = z^2 - 2z + 2$
is a factor of $\displaystyle z^4+3z^2-6z+10$. So do the division. This gives you a quadratic in z to solve, which you can do by factoring or the quadratic formula or whatnot.
-Dan