If z = (2 – 3i), prove that z2 – 4z + 13 = 0 and hence deduce that 4z3 – 3z2 + 169 = 0.
Given: z = 2 – 3i
To Prove: z2 – 4z + 13 = 0
Taking LHS, z2 – 4z + 13
Putting the value of z = 2 – 3i, we get
(2 – 3i)2 – 4(2 – 3i) + 13
= 4 + 9i2 – 12i – 8 + 12i + 13
= 9(-1) + 9
= - 9 + 9
= 0
= RHS
Hence, z2 – 4z + 13 = 0 …(i)
Now, we have to deduce 4z3 – 3z2 + 169
Now, we will expand 4z3 – 3z2 + 169 in this way so that we can use the above equation i.e. z2 – 4z + 13
= 4z3 – 16z2 + 13z2 +52z – 52z + 169
Re – arrange the terms,
= 4z3 – 16z2 + 52z + 13z2 – 52z + 169
= 4z(z2 – 4z + 13) + 13 (z2 – 4z + 13)
= 4z(0) + 13(0) [from eq. (i)]
= 0
= RHS
Hence Proved