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

Step-by-step explanation:

We have,

$z = 3 - 2i$

$\implies \: z - 3 = - 2i$

$\implies \:( z - 3)^{2} = (- 2i) ^{2}$

$\implies \:z ^{2} + {3}^{2} - 6z = (- 2) ^{2}. i ^{2}$

$\implies \:z ^{2} - 6z + 9= 4.( - 1)$

$\implies \:z ^{2} - 6z + 9= - 4$

$\implies \:z ^{2} - 6z + 9 + 4 = 0$

$\implies \:z ^{2} - 6z + 13 = 0$

Now, we have,

$z^{4}-4z^{3}+6z^{2}-4z+17$

$=z^{4}-6z^{3}+13z^{2}+2z^{3}-7z^{2}-4z+17$

$=z^{2}(z^{2}-6z+13)+2z^{3}-7z^{2}-4z+17$

$=z^{2}(z^{2}-6z+13)+2z^{3}-12z^{2}+26z+5z^{2}-30z+17$

$=z^{2}(z^{2}-6z+13)+2z(z^{2}-6z+13z)+5z^{2}-30z+65-48$

$=z^{2}(z^{2}-6z+13)+2z(z^{2}-6z+13z)+5(z^{2}-6z+13)-48$

Since, $z^{2}-6z+13=0$, so,

$=z^{2}(0)+2z(0)+5(0)-48$

$=0+0+0-48$

$= -48$