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If z= 3 – 5i , then show that z3 – 10z2 + 58z – 136 = 0 |
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Answer» z = 3 – 5i ................. (1) ⇒ z2 = (3 – 5i)2 = (3)2 – 2(3) (5i) + (5i)2 = 9 – 30i + 25i2 = 9 – 30i + 25 (–1) = – 16 – 30i ∴ z2 = – 16 – 30i ................. (2) Now z3 = z2 .z = (–16 – 30i) (3 – 5i) = – 48 – 90i + 80i + 150i2 = – 48 – 10i + 150 (–1) = – 198 – 10i ∴ z3 = – 198 – 10i ................. (3) Now z3 – 10z2 + 58z – 136 = (–198 – 10i) – 10 (–16 – 30i) + 58 (3 – 5i) – 136 = – 198 – 10i + 160 + 300i + 174 – 290i – 136 = (–198 + 160 + 174 – 136) + i(–10 + 300 – 290) = 0 + i(0) = 0 ∴ z3 – 10z2 + 58z – 136 = 0 |
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