4-1.Complex numbers
hard

જો $z = 3 - 4i$, તો ${z^4} - 3{z^3} + 3{z^2} + 99z - 95 =$ . . .

A

$5$

B

$6$

C

$-5$

D

$-4$

Solution

(a)Given that $z = 3 – 4i$==> ${z^2} = – 7 – 24i$,
${z^4} = – 117 – 44i$ and ${z^4} = – 527 + 336i$
${z^4} – 3{z^3} + 3{z^2} + 99z – 95 = 5$
Aliter : $z = 3 – 4i$==> ${(z – 3)^2} = – 16$
==>${z^2} – 6z + 25 = 0$
${z^4} – 3{z^3} + 3{z^2} + 99z – 95$
$ = ({z^2} + 3z – 4)({z^2} – 6z + 25) + 5$$ = ({z^2} + 3z – 4)(0) + 5 = 5$

Standard 11
Mathematics

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