If `z=2-3i`show that `z^2-4z+13=0`and hence find the value of `4z^3-3z

1.	If `z=2-3i`show that `z^2-4z+13=0`and hence find the value of `4z^3-3z^2+169.`
Answer» `z = 2-3i` Now, `z^2-4z+13 = (2-3i)^2-4(2-3i)+13` `=4+9i^2-12i-8+12i+13` `=4-9+5 = 0...[As i^2 = -1]` `:. z^2-4z+13 = 0` Now, `4z^3-3z^2+169 = 4z^3-16z^2+13z^2+52z-52z+169` `=4z^3-16z^2+52z+13z^2-52z+169` `=4z(z^2-4z+13) + 13(z^2-4z+13) ` `=4z(0)+13(0) = 0` `:. 4z^3-3z^2+169 = 0.`

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If `z=2-3i`show that `z^2-4z+13=0`and hence find the value of `4z^3-3z^2+169.`

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