# 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

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