Question

if z=2-1 prove that
${z}^{3} + 3 {z}^{2} - 9z + 8 = 1 + 14i$

Answer 1

We have,

$z = 2 + i \: \: \: \: .......(1)\\ \\ z - 2 = i$
On squaring both sides
$\\ \\ {(z - 2)}^{2} = {i}^{2} \\ \\ (since \: \: \: \: {i}^{2} = - 1) \\ \\ {z}^{2} + 4 - 4z = - 1 \\ \\ {z}^{2} - 4z + 5 = 0 \: \: \: \: \: .......(2)$

$\\ LHS = {z}^{3} + 3 {z}^{2} - 9z + 8 \\ \\ = {z}^{3} - 4 {z}^{2} + 5z + 7 {z}^{2} - 28z + 35 + 14z - 27 \\ \\ = z( {z}^{2} - 4z + 5) + 7( {z}^{2} - 4z + 5) + 14z - 27 \\ \\ using \: eq(2) \\ \\ = z \times 0 + 7 \times 0 + 14z - 27 \\ \\ = 14z - 27 \\ \\ using \: eq \: (2) \\ \\ = 14(2 + i) - 27 \\ \\ = 28 + 14i - 27 \\ \\ = 1 + 14i \\ \\ = RHS$

HENCE PROVED

⭐ hope it may help you⭐

Answer 2

yes he is right........