If z = 3 - 4i,then {z^4} - 3{z^3} + 3{z^2} + | vedclass

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(A) Given $z = 3 - 4i$,we have $z - 3 = -4i$. Squaring both sides,we get $(z - 3)^2 = (-4i)^2$. $z^2 - 6z + 9 = 16i^2$. Since $i^2 = -1$,we have $z^2

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$5$

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(A) Given $z = 3 - 4i$,we have $z - 3 = -4i$. Squaring both sides,we get $(z - 3)^2 = (-4i)^2$. $z^2 - 6z + 9 = 16i^2$. Since $i^2 = -1$,we have $z^2 - 6z + 9 = -16$,which simplifies to $z^2 - 6z + 25 = 0$. Now,we perform polynomial division of $P(z) = z^4 - 3z^3 + 3z^2 + 99z - 95$ by $z^2 - 6z + 25$. $z^4 - 3z^3 + 3z^2 + 99z - 95 = (z^2 + 3z - 4)(z^2 - 6z + 25) + 5$. Substituting $z^2 - 6z + 25 = 0$,we get: $P(z) = (z^2 + 3z - 4)(0) + 5 = 5$.

If $z = 3 - 4i$,then ${z^4} - 3{z^3} + 3{z^2} + 99z - 95$ is equal to

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