Q. 244.0( 4 Votes )

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


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