1. **State the problem:** Given two complex numbers $z_1 = 5 + 3i$ and $z_2 = 2 - i$, (a) find their conjugates $\overline{z_1}$ and $\overline{z_2}$. (b) Determine the value of $k$ if $\frac{1}{z_2} = k \overline{z_1}$. 2. **Recall the formula for conjugate:** The conjugate of a complex number $a + bi$ is $a - bi$. 3. **Find the conjugates:** $$\overline{z_1} = 5 - 3i$$ $$\overline{z_2} = 2 + i$$ 4. **Express $\frac{1}{z_2}$:** To find $\frac{1}{z_2}$, multiply numerator and denominator by the conjugate of $z_2$: $$\frac{1}{2 - i} \times \frac{2 + i}{2 + i} = \frac{2 + i}{(2)^2 - (-1)^2} = \frac{2 + i}{4 + 1} = \frac{2 + i}{5}$$ 5. **Set up the equation:** $$\frac{1}{z_2} = k \overline{z_1} \implies \frac{2 + i}{5} = k (5 - 3i)$$ 6. **Solve for $k$:** Divide both sides by $5 - 3i$: $$k = \frac{\frac{2 + i}{5}}{5 - 3i} = \frac{2 + i}{5} \times \frac{1}{5 - 3i}$$ Multiply numerator and denominator by the conjugate of the denominator: $$k = \frac{2 + i}{5} \times \frac{5 + 3i}{(5)^2 - (-3)^2} = \frac{2 + i}{5} \times \frac{5 + 3i}{25 + 9} = \frac{2 + i}{5} \times \frac{5 + 3i}{34}$$ 7. **Multiply the numerators:** $$(2 + i)(5 + 3i) = 2 \times 5 + 2 \times 3i + i \times 5 + i \times 3i = 10 + 6i + 5i + 3i^2 = 10 + 11i + 3(-1) = 10 + 11i - 3 = 7 + 11i$$ 8. **Combine all:** $$k = \frac{7 + 11i}{5 \times 34} = \frac{7 + 11i}{170}$$ **Final answer:** $$\overline{z_1} = 5 - 3i$$ $$\overline{z_2} = 2 + i$$ $$k = \frac{7 + 11i}{170}$$