Determine a third-degree polynomial with real coefficients…
Determine a third-degree polynomial with real coefficients that satisfies the conditions: $$ P(1-i)=2+i \qquad \text{and} \qquad P(i)=1-2i. $$
Explanation of Method
A cubic $P(x)=ax^3+bx^2+cx+d$ has four real unknowns, so two complex conditions — each splitting into a real and an imaginary equation — give exactly the four real equations needed to solve for $a,b,c,d$. Since $P$ has real coefficients, it automatically satisfies $P(\bar z)=\overline{P(z)}$, so the two given conditions also determine $P(1+i)$ and $P(-i)$ for free, which is useful for verification.
Step 1 · Note the conjugate consequences
Since $P$ has real coefficients, $P(1-i)=2+i$ forces $P(1+i)=2-i$, and $P(i)=1-2i$ forces $P(-i)=1+2i$. These will serve as a check later.
Step 2 · Compute powers of $1-i$
$$ (1-i)^2 = 1-2i+i^2 = -2i, \qquad (1-i)^3 = (1-i)(-2i) = -2i+2i^2 = -2-2i $$
Step 3 · Evaluate $P(1-i)$
$$ P(1-i) = a(-2-2i)+b(-2i)+c(1-i)+d = (-2a+c+d) + i(-2a-2b-c) $$
Setting this equal to $2+i$:
$$ -2a+c+d = 2 \qquad (5) \qquad\qquad -2a-2b-c = 1 \qquad (6) $$
Step 4 · Evaluate $P(i)$
Using $i^2=-1$, $i^3=-i$:
$$ P(i) = -ai-b+ci+d = (-b+d)+i(-a+c) $$
Setting this equal to $1-2i$:
$$ -b+d = 1 \qquad (7) \qquad\qquad -a+c = -2 \qquad (8) $$
Step 5 · Solve the system
From (8): $c=a-2$. Substituting into (6): $-2a-2b-(a-2)=1 \Rightarrow -3a-2b=-1 \Rightarrow 3a+2b=1$ (9).
From (7): $d=b+1$. Substituting $c=a-2$, $d=b+1$ into (5): $-2a+(a-2)+(b+1)=2 \Rightarrow -a+b=3 \Rightarrow b=a+3$ (10).
Substituting (10) into (9): $3a+2(a+3)=1 \Rightarrow 5a+6=1 \Rightarrow 5a=-5 \Rightarrow a=-1$.
Then $b=a+3=2$, $c=a-2=-3$, $d=b+1=3$.
$$ \boxed{a=-1,\ b=2,\ c=-3,\ d=3} \qquad\Longrightarrow\qquad P(x) = -x^3+2x^2-3x+3 $$
Step 6 · Verify both given conditions
$$ P(1-i) = -(-2-2i)+2(-2i)-3(1-i)+3 = (2+2i)-4i-3+3i+3 = (2-3+3)+i(2-4+3) = 2+i \quad\checkmark $$
$$ P(i) = -(-i)+2(-1)-3i+3 = i-2-3i+3 = (-2+3)+i(1-3) = 1-2i \quad\checkmark $$
Step 7 · Verify the conjugate consequences
Using $(1+i)^2=2i$, $(1+i)^3=(1+i)(2i)=-2+2i$:
$$ P(1+i) = -(-2+2i)+2(2i)-3(1+i)+3 = (2-2i)+4i-3-3i+3 = (2-3+3)+i(-2+4-3) = 2-i \quad\checkmark $$
$$ P(-i) = -(-i)^3+2(-i)^2-3(-i)+3 = -i-2+3i+3 = 1+2i \quad\checkmark $$
Both conjugate conditions hold automatically, confirming the solution.
Step 8 · Uniqueness
The two complex conditions split into exactly four independent real linear equations in the four unknowns $a,b,c,d$, and the resulting $4\times4$ linear system has a unique solution — so this is the only cubic with real coefficients meeting both requirements.
Result
Verified against both given conditions and, via the real-coefficient conjugate property $P(\bar z)=\overline{P(z)}$, against the two implied conditions $P(1+i)=2-i$ and $P(-i)=1+2i$ as well.