Question

Find the roots of given equation:

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

• A. $1,1+\sqrt{3}i,1-\sqrt{3}i$
• B. $1,1+\sqrt{2}i,1-\sqrt{2}i$
• C. $3,1+\sqrt{3}i,1-\sqrt{3}i$
• D. All of them

Answer 1

The correct option is A $1,1+\sqrt{3}i,1-\sqrt{3}i$

In this case, for

$p\left(z\right)=z^3-3z^2+6z-4$,

we have that $p\left(1\right)=0$.

Therefore, we can factorize it further and get

$z^3-3z^2+6z-4=(z-1)(z^2-2z+4)$

$=(z-1)\left(\right(z-1{)}^2+3)$.

Their roots are just

$z_1=1,z_2=1+i\sqrt{3},z_3=1-i\sqrt{3}$.