Complex numbers-proving and finding roots

**kingkaisai2** · July 31st 2006, 01:23 AM

Prove that 1+i is a root of the equation z^4+3z^2-6z+10=0. Find all the other roots.

**topsquark** · July 31st 2006, 04:25 AM

Originally Posted by kingkaisai2

Prove that 1+i is a root of the equation z^4+3z^2-6z+10=0. Find all the other roots.

$z^4+3z^2-6z+10=0$

$(1+i)^4+3(1+i)^2-6(1+i)+10$

$(1^4+4*1^3*i+6*1^2*i^2+4*1*i^3+i^4)$ $+3(1^2+2*1*i+i^2)-6(1+i)+10$

$(1 + 4i - 6 - 4i + 1)+ 3(1 + 2i - 1) - 6(1 + i) + 10$

$-4 + 6i - 6 - 6i + 10$

$0$ (Check.)

Now, if 1+i is a root, so is 1-i. Thus we know that $\left ( z - (1+i) \right ) \left ( z - (1-i) \right )$ $= \left ( (z-1) - i \right ) \left ( (z-1) + i \right )$ $= (z-1)^2 - i^2 = z^2 - 2z + 1 + 1 = z^2 - 2z + 2$
is a factor of $z^4+3z^2-6z+10$ . 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

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