If z_1 and z_2 are solutions of z^2-4z+13=0 , show that…
If $z_1$ and $z_2$ are solutions of $z^2-4z+13=0$, show that: $$ 1^{\circ}\,\, \overline{(z_1-z_2)(z_2+4)} = (z_2-z_1)(z_1+4); \qquad 2^{\circ}\,\, \overline{2z_2^2-5i} = 2z_1^2+5i; \qquad 3^{\circ}\,\, \overline{\left(\frac{z_2}{z_1}\right)} = \frac{z_1}{z_2}. $$
Explanation of Method
Since the equation has real coefficients, its two roots are complex conjugates: $z_2=\bar z_1$ and $z_1=\bar z_2$. Consequently, for any expression $F(z_1,z_2,i)$ built from $z_1$, $z_2$, real numbers, and $i$, taking the conjugate of the whole expression is the same as swapping $z_1\leftrightarrow z_2$ and replacing every $i$ by $-i$. All three identities below are instances of this single principle.
Preliminary · Find the roots
$$ z = \frac{4\pm\sqrt{16-52}}{2} = \frac{4\pm\sqrt{-36}}{2} = \frac{4\pm6i}{2} = 2\pm3i $$
$$ z_1 = 2+3i, \qquad z_2 = 2-3i $$
Since the coefficients $1,-4,13$ are real, $z_2=\bar z_1$ and $z_1=\bar z_2$. By Vieta's formulas: $z_1+z_2=4$, $z_1z_2=13$.
Part $1^{\circ}$ · $\overline{(z_1-z_2)(z_2+4)} = (z_2-z_1)(z_1+4)$
Step 1 · Conjugate the product
$$ \overline{(z_1-z_2)(z_2+4)} = \overline{(z_1-z_2)}\cdot\overline{(z_2+4)} $$
Step 2 · Apply $\bar z_1=z_2$, $\bar z_2=z_1$
$$ \overline{z_1-z_2} = \bar z_1-\bar z_2 = z_2-z_1, \qquad \overline{z_2+4} = \bar z_2+4 = z_1+4 $$
Step 3 · Combine
$$ \boxed{\overline{(z_1-z_2)(z_2+4)} = (z_2-z_1)(z_1+4)} $$
Step 4 · Numerical check
$$ (z_1-z_2)(z_2+4) = (6i)(6-3i) = 36i-18i^2 = 18+36i \quad\Longrightarrow\quad \overline{(z_1-z_2)(z_2+4)} = 18-36i $$
$$ (z_2-z_1)(z_1+4) = (-6i)(6+3i) = -36i-18i^2 = 18-36i \quad \checkmark $$
Part $2^{\circ}$ · $\overline{2z_2^2-5i} = 2z_1^2+5i$
Step 1 · Distribute the conjugate
$$ \overline{2z_2^2-5i} = 2\overline{z_2^2} - \overline{5i} = 2\overline{z_2^2}+5i $$
using $\overline{-5i}=5i$.
Step 2 · Use $\bar z_2=z_1$
$$ \overline{z_2^2} = \overline{z_2}\,^2 = z_1^2 $$
Step 3 · Conclude
$$ \boxed{\overline{2z_2^2-5i} = 2z_1^2+5i} $$
Step 4 · Numerical check
$$ z_2^2 = (2-3i)^2 = 4-12i-9 = -5-12i \quad\Longrightarrow\quad 2z_2^2-5i = -10-24i-5i = -10-29i $$
$$ \overline{2z_2^2-5i} = -10+29i $$
$$ z_1^2 = (2+3i)^2 = -5+12i \quad\Longrightarrow\quad 2z_1^2+5i = -10+24i+5i = -10+29i \quad \checkmark $$
Part $3^{\circ}$ · $\overline{\left(\dfrac{z_2}{z_1}\right)} = \dfrac{z_1}{z_2}$
Step 1 · Conjugate a quotient
$$ \overline{\left(\frac{z_2}{z_1}\right)} = \frac{\bar z_2}{\bar z_1} $$
Step 2 · Apply $\bar z_2=z_1$, $\bar z_1=z_2$
$$ \boxed{\overline{\left(\frac{z_2}{z_1}\right)} = \frac{z_1}{z_2}} $$
Step 3 · Numerical check
$$ \frac{z_2}{z_1} = \frac{2-3i}{2+3i}\cdot\frac{2-3i}{2-3i} = \frac{(2-3i)^2}{13} = \frac{-5-12i}{13} \quad\Longrightarrow\quad \overline{\left(\frac{z_2}{z_1}\right)} = \frac{-5+12i}{13} $$
$$ \frac{z_1}{z_2} = \frac{2+3i}{2-3i}\cdot\frac{2+3i}{2+3i} = \frac{(2+3i)^2}{13} = \frac{-5+12i}{13} \quad \checkmark $$
| Part | Identity | Underlying rule |
|---|---|---|
| $1^{\circ}$ | $\overline{(z_1-z_2)(z_2+4)}=(z_2-z_1)(z_1+4)$ | Swap $z_1\leftrightarrow z_2$ |
| $2^{\circ}$ | $\overline{2z_2^2-5i}=2z_1^2+5i$ | Swap $z_1\leftrightarrow z_2$, flip sign of $i$ |
| $3^{\circ}$ | $\overline{z_2/z_1}=z_1/z_2$ | Swap $z_1\leftrightarrow z_2$ |
Result
All three identities are consequences of a single fact: because $z_1,z_2$ are roots of a real-coefficient quadratic, $\bar z_1=z_2$ and $\bar z_2=z_1$. Conjugating any polynomial or rational expression in $z_1$, $z_2$, and $i$ therefore amounts to swapping $z_1\leftrightarrow z_2$ and negating every explicit $i$.