If z = 2 − i√7 , then show that 3z3 – 4z2 + z + 88 = 0.

1.	If z = 2 − i√7 , then show that 3z3 – 4z2 + z + 88 = 0.
Answer» Given z = 2− i√7 ⇒ z – 2 = − i√7 ⇒ (z – 2)² = (− i√7)² ⇒ z² – 4z + 4 = 7i² ⇒ z² – 4z + 4 = – 7 ⇒ z² – 4z + 11 = 0 3z³ –4z² + z + 88 = 3z(z² – 4z + 11) + 8z² – 32z + 88 = 3z(0) + 8(z² – 4z + 11) = 0 + 8(0) = 0 ∴ 3z³ – 4z² + z + 88 = 0

Answer»

Given z = 2− i√7

⇒ z – 2 = − i√7

⇒ (z – 2)² = (− i√7)²

⇒ z² – 4z + 4 = 7i²

⇒ z² – 4z + 4 = – 7

⇒ z² – 4z + 11 = 0

3z³ –4z² + z + 88 = 3z(z² – 4z + 11) + 8z² – 32z + 88

= 3z(0) + 8(z² – 4z + 11) = 0 + 8(0) = 0

∴ 3z³ – 4z² + z + 88 = 0

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If z = 2 − i√7 , then show that 3z3 – 4z2 + z + 88 = 0.

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