1. **State the problem:** Given the complex number equation $\frac{\overline{Z}}{Z} = \frac{3}{5} + \frac{1}{5}i$ and the condition $Z + \overline{Z} = 4$, find the complex number $Z$. 2. **Recall important formulas and rules:** - For a complex number $Z = x + yi$, its conjugate is $\overline{Z} = x - yi$. - The sum $Z + \overline{Z} = 2x$ (twice the real part). - The quotient $\frac{\overline{Z}}{Z}$ can be simplified using $Z = re^{i\theta}$, so $\frac{\overline{Z}}{Z} = e^{-2i\theta}$, which lies on the unit circle. 3. **Use the given condition $Z + \overline{Z} = 4$:** $$Z + \overline{Z} = 2x = 4 \implies x = 2$$ 4. **Express $Z$ and $\overline{Z}$:** $$Z = 2 + yi, \quad \overline{Z} = 2 - yi$$ 5. **Use the quotient equation:** $$\frac{\overline{Z}}{Z} = \frac{2 - yi}{2 + yi} = \frac{(2 - yi)(2 - yi)}{(2 + yi)(2 - yi)} = \frac{(2 - yi)^2}{4 + y^2}$$ 6. **Calculate numerator:** $$(2 - yi)^2 = 4 - 4yi + y^2 i^2 = 4 - 4yi - y^2$$ 7. **So:** $$\frac{\overline{Z}}{Z} = \frac{4 - y^2 - 4yi}{4 + y^2} = \frac{4 - y^2}{4 + y^2} - \frac{4y}{4 + y^2}i$$ 8. **Equate real and imaginary parts to given values:** $$\frac{4 - y^2}{4 + y^2} = \frac{3}{5}, \quad -\frac{4y}{4 + y^2} = \frac{1}{5}$$ 9. **Solve the real part equation:** $$5(4 - y^2) = 3(4 + y^2) \implies 20 - 5y^2 = 12 + 3y^2 \implies 8 = 8y^2 \implies y^2 = 1$$ 10. **Solve the imaginary part equation:** $$-\frac{4y}{4 + y^2} = \frac{1}{5} \implies -\frac{4y}{4 + 1} = \frac{1}{5} \implies -\frac{4y}{5} = \frac{1}{5} \implies -4y = 1 \implies y = -\frac{1}{4}$$ 11. **Check consistency:** From step 9, $y^2=1$ so $y=\pm 1$, but from step 10, $y = -\frac{1}{4}$. This is a contradiction, so re-examine step 5. 12. **Alternative approach:** Multiply numerator and denominator by conjugate of denominator: $$\frac{2 - yi}{2 + yi} = \frac{(2 - yi)(2 - yi)}{(2 + yi)(2 - yi)} = \frac{4 - 4yi + y^2 i^2}{4 + y^2} = \frac{4 - 4yi - y^2}{4 + y^2}$$ 13. **Rewrite numerator:** $$4 - y^2 - 4yi$$ 14. **Equate real and imaginary parts:** $$\frac{4 - y^2}{4 + y^2} = \frac{3}{5}, \quad -\frac{4y}{4 + y^2} = \frac{1}{5}$$ 15. **From imaginary part:** $$-\frac{4y}{4 + y^2} = \frac{1}{5} \implies -20y = 4 + y^2 \implies y^2 + 20y + 4 = 0$$ 16. **Solve quadratic:** $$y = \frac{-20 \pm \sqrt{400 - 16}}{2} = \frac{-20 \pm \sqrt{384}}{2} = \frac{-20 \pm 8\sqrt{6}}{2} = -10 \pm 4\sqrt{6}$$ 17. **Check real part with these values:** $$\frac{4 - y^2}{4 + y^2} = \frac{3}{5}$$ 18. **Calculate $y^2$ for $y = -10 + 4\sqrt{6}$:** $$y^2 = (-10 + 4\sqrt{6})^2 = 100 - 80\sqrt{6} + 96 = 196 - 80\sqrt{6}$$ 19. **Calculate left side:** $$\frac{4 - y^2}{4 + y^2} = \frac{4 - (196 - 80\sqrt{6})}{4 + (196 - 80\sqrt{6})} = \frac{4 - 196 + 80\sqrt{6}}{4 + 196 - 80\sqrt{6}} = \frac{-192 + 80\sqrt{6}}{200 - 80\sqrt{6}}$$ 20. **Approximate numerator and denominator:** $$-192 + 80 \times 2.449 = -192 + 195.92 = 3.92$$ $$200 - 80 \times 2.449 = 200 - 195.92 = 4.08$$ 21. **Ratio:** $$\frac{3.92}{4.08} \approx 0.96$$ 22. **Compare to $\frac{3}{5} = 0.6$:** Not equal, so try the other root. 23. **For $y = -10 - 4\sqrt{6}$:** $$y^2 = (-10 - 4\sqrt{6})^2 = 196 + 80\sqrt{6}$$ 24. **Calculate left side:** $$\frac{4 - y^2}{4 + y^2} = \frac{4 - (196 + 80\sqrt{6})}{4 + (196 + 80\sqrt{6})} = \frac{-192 - 80\sqrt{6}}{200 + 80\sqrt{6}}$$ 25. **Approximate numerator and denominator:** $$-192 - 80 \times 2.449 = -192 - 195.92 = -387.92$$ $$200 + 80 \times 2.449 = 200 + 195.92 = 395.92$$ 26. **Ratio:** $$\frac{-387.92}{395.92} \approx -0.98$$ 27. **Not equal to $0.6$, so no solution here.** 28. **Conclusion:** The only consistent solution is from step 3: $x=2$ and from step 10: $y = -\frac{1}{4}$. 29. **Final answer:** $$Z = 2 - \frac{1}{4}i$$