Question

if z=3-2i then show that z^2-6c+13=0 hence find the value of z^4 - 4z^3+6z^2-4z+17​

Answer 1

Answer:

-48

Step-by-step explanation:

Given,

z =3-2i

L.s= z^2-6z+13

=(3-2i)^2 - 6(3-2i) +13

= 9-12i+4i^2-18+12i+13

=9-4-18+13

=22-22

=0

=R.S. (Showed)

Given,

z^4-4z^3+6z^2-4z+17= z(z^3-4z^2+6z-4)+17

${z}^{4} - 4 {z}^{3} + 6 {z}^{2} - 4z + 17 \\ = z( {z}^{3} - 2 {z}^{2} - 2 {z}^{2} + 4z + 2z - 4) \\ + 17$

=z{z^2(z-2)-2z(z-2)+2(z-2)}+17

=z{(z-2)(z^2-2z+2)} +17

=(z^2-2z) (z^2-2z+2) +17

Now,

z^2-2z =(3-2i)^2 -2(3-2i) =9-12i-10+4i=-1-8i

and, z^2-2z+2 = -1+2-8i = 1-8i

so, (z^2-2z) (z^2-2z +2)=-(1+8i)(1-8i)=-(1+64)=-65

So, z^4-4z^3+6z^2-4z+17=-65+17=-48