Question

If z= 3 – 5i , then show that z3 – 10z2 + 58z – 136 = 0

Answer 1

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