If f(z) = 2 + z + 3z^2 , find f(z) and f(bar z) for z = 3+2i
If $f(z) = 2 + z + 3z^2$, find $f(z)$ and $f(\bar z)$ for $z = 3+2i$.
Explanation of Method
Substitute $z=3+2i$ directly into the polynomial, computing $z^2$ first via $(a+bi)^2 = a^2-b^2+2abi$. For $f(\bar z)$, either substitute $\bar z = 3-2i$ directly, or use the fact that a polynomial with real coefficients satisfies $f(\bar z) = \overline{f(z)}$ — computing one value then immediately gives the other by conjugation.
Step 1 · Identify $z$ and $\bar z$
$$ z = 3+2i, \qquad \bar z = \overline{3+2i} = 3-2i $$
Step 2 · Compute $z^2$
$$ z^2 = (3+2i)^2 = 3^2 + 2(3)(2i) + (2i)^2 = 9+12i+4i^2 = 9+12i-4 = 13+12i $$
Step 3 · Compute $f(z)$
$$ f(z) = 2+z+3z^2 = 2+(3+2i)+3(13+12i) = 2+3+2i+39+36i $$
$$ = (2+3+39) + (2+36)i = \boxed{44+38i} $$
Step 4 · Compute $\bar z^2$
$$ \bar z^2 = (3-2i)^2 = 9-12i+4i^2 = 9-12i-4 = 13-12i $$
As expected, $\bar z^2 = \overline{z^2}$ (conjugating $z$ before squaring gives the same result as squaring then conjugating).
Step 5 · Compute $f(\bar z)$
$$ f(\bar z) = 2+\bar z+3\bar z^2 = 2+(3-2i)+3(13-12i) = 2+3-2i+39-36i $$
$$ = (2+3+39) + (-2-36)i = \boxed{44-38i} $$
Step 6 · Why $f(\bar z) = \overline{f(z)}$
Since $f$ has real coefficients ($2$ and $3$), conjugation distributes through the polynomial:
$$ f(\bar z) = 2+\bar z+3\bar z^2 = \overline{2}+\overline{z}+\overline{3z^2} = \overline{2+z+3z^2} = \overline{f(z)} $$
and indeed $44-38i = \overline{44+38i}$, confirming the two computations agree. ✓
Step 7 · Additional observations
$\text{Re}(f(z)) = \text{Re}(f(\bar z)) = 44$, and $\text{Im}(f(z)) = -\text{Im}(f(\bar z)) = 38$.
$$ |f(z)| = |f(\bar z)| = \sqrt{44^2+38^2} = \sqrt{1936+1444} = \sqrt{3380} = 26\sqrt5 $$
$z$ and $\bar z$ are mirror images across the real axis, and so are $f(z)$ and $f(\bar z)$ — a consequence of $f$ having real coefficients (not to scale).
Result
As guaranteed for any polynomial with real coefficients, $f(\bar z) = \overline{f(z)}$.