- Home
- Standard 11
- Mathematics
4-1.Complex numbers
hard
If $z = 3 - 4i$, then ${z^4} - 3{z^3} + 3{z^2} + 99z - 95$ is equal to
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
