1. Consider the polynomial \(p(x) = x^3 + (1-i)x^2 + 2i x + 3+4i.\)Calculate \(p(1), p(i), p(1+i), p(1-i)\).
2. Consider the polynomial \(p(z) = z^3 +2z^2 - z + 4.\)Calculate \(p(2+i), p(2-i), p(1+3i), p(1-3i)\).
3. Let p be a polynomial with real coefficients such that \(p(1-2i) = 1+3i \text{ and } p(3+4i) = 2 - i.\)Calculate \(p(1 +2i), p(2+4i), p(3-4i)\)if any of these are not uniquely determined, enter ? for that value.
Could someone help me with these please? Thanks!
+128460 1. Remember that i^2 = -1 i^3 = -i i^4 =1
P(1) = (1)^3 + (1 - i)(1)^2 + 2i(1) + 3 +4i
= 1 + 1 - i + 2i + 3 + 4i
= 5 + 5i
P(i) = (i)^3 + (1 -i)(i)^2 + 2i(i) + 3 + 4i
= -i + (1 - i)(-1) + 2i^2 + 3 + 4i
= -i - 1 + i + 2(-1) + 3 + 4i
= -1 - 2 + 3 + 4i
= 4i
P(1 + i) = (1 + i)^3 + (1 - i) (1 + i)^2 + 2i (1 + i) + 3 + 4i
Note that we can write
( 1 + i) [ (1 + i)^2 + (1 - i) ( (1 + i) + 2i ] +3 + 4i =
(1 + i) [ (1 + 2i +i^2 + 1 - i^2 + 2i ] + 3 + 4i =
(1 + i) [ 1 + 2i - 1 + 1 + 1 + 2i ] + 3 + 4i =
(1 + i) [ 2 + 4i) + 3 + 4i =
(2 + 2i + 4i + 4i^2) + 3 + 4i =
(2 + 6i - 4 ) + 3 + 4i =
(6i - 2) + 3 + 4i =
1 + 10i
P(1 - i) = (1 - i)^3 + (1 - i)(1 - i)^2 + 2i (1 - i) + 3 + 4i
And we can write
(1 - i) [ (1 - i)^2 +(1 - i)^2 + 2i ] + 3 + 4i =
( 1 - i) [ 1 - 2i + i^2 + 1 - 2i + i^2 + 2i ] + 3 + 4i =
(1 - i) [ [ 1 - 2i - 1 + 1 -2i -1 + 2i ] + 3 + 4i =
(1 - i) [ -2i] + 3 + 4i =
-2i + 2i^2 + 3 + 4i =
-2i - 2 + 3 + 4i =
1 + 2i
+128460 2.
p(2 + i) = (2 + i)^3 + 2(2 + i)^2 - (2 + i) + 4
= (2 + i ) [ (2 + i)^2 + 2 (2 + i) - 1 ] + 4
= (2 + i) [ 4 + 4i + i^2 + 4 + 2i - 1 ] + 4
= (2 + i) [ 4 + 4i - 1 + 4 + 2i - 1 ]+ 4
= (2 + i) [ 6 + 6i ] + 4
= 12 + 6i + 12i + 6i^2 + 4
= 12 + 18i - 6 + 4
= 10 + 18i
P(2 - i) = (2 - i)^3 + 2(2 - i)^2 - (2 - i) + 4
= (2 -i) [ (2 - i)^2 + 2(2 -i) - 1 ] + 4
= (2 -i) [ 4 - 4i + i^2 + 4 - 2i - 1 ] + 4
= ( 2 -i) [ [ 4 - 4i - 1 + 4 - 2i - 1 ] + 4
= ( 2 - i) [ 6 - 6i ] + 4
= 12 - 6i - 12i + 6i^2 + 4
= 12 - 6i - 12i - 6 + 4
= 10 - 18i
P(1 + 3i) = (1 + 3i)^3 + 2(1 + 3i)^2 - (1 + 3i) + 4
= ( 1 + 3i) [ ( 1 + 3i)^2 + 2(1 + 3i) - 1 ] + 4
= (1 + 3i) [ 1 + 6i + 9i^2 + 2 + 6i - 1 ] + 4
= (1 + 3i) [ 1 + 6i - 9 + 2 +6i - 1 ]+ 4
= ( 1 + 3i) [ 12i -7 ] + 4
= ( 12i + 36i^2 - 7 - 21i) + 4
= ( -9i - 36 - 7 ] + 4
= - 39 - 9i
Note that P(2 + i) and P(2 - i) only differed by signs on the "i" term
So....we would expect that P( 1 - 3i) would just be -39 + 9i
