If z = (2 – 3i), prove that z2 – 4z + 13 = 0 and hence deduce that

1.	If z = (2 – 3i), prove that z2 – 4z + 13 = 0 and hence deduce that 4z3 – 3z2 + 169 = 0.
Answer» 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

If z = (2 – 3i), prove that z2 – 4z + 13 = 0 and hence deduce that 4z3 – 3z2 + 169 = 0.

Answer»

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