1. **Problem (a):** Given $f'(x) = \frac{1}{x}$ and $f(x) = \int_1^x \frac{1}{t} dt$, show the relationship. 2. The derivative $f'(x) = \frac{1}{x}$ is the derivative of the natural logarithm function $\ln|x|$. 3. By the Fundamental Theorem of Calculus, $f(x) = \int_1^x \frac{1}{t} dt = \ln|x| - \ln 1 = \ln x$ for $x > 0$. 4. **Problem (b):** Show $x - y = 1$ if and only if $\frac{e^y}{e^x} = e^{-1}$. 5. Starting from $x - y = 1$, rearranged as $y = x - 1$. 6. Then $\frac{e^y}{e^x} = e^{y - x} = e^{(x - 1) - x} = e^{-1}$. 7. Conversely, if $\frac{e^y}{e^x} = e^{-1}$, then $e^{y - x} = e^{-1}$ implies $y - x = -1$ or $x - y = 1$. 8. **Problem (c):** Given complex numbers $z_1 = 2 + i$, $z_2 = -2 + 4i$, and $z_3 = \frac{2 z_2}{z_1}$, show $\arg(z_3) = 2 \arg(z_2) - \arg(z_1) + \frac{\pi}{2}$. 9. Calculate $z_3$: $$z_3 = \frac{2(-2 + 4i)}{2 + i} = \frac{-4 + 8i}{2 + i}.$$ 10. Multiply numerator and denominator by conjugate of denominator: $$z_3 = \frac{(-4 + 8i)(2 - i)}{(2 + i)(2 - i)} = \frac{-8 + 4i + 16i - 8i^2}{4 + 1} = \frac{-8 + 20i + 8}{5} = \frac{0 + 20i}{5} = 4i.$$ 11. So $z_3 = 4i$. 12. Calculate arguments: - $\arg(z_1) = \tan^{-1}(\frac{1}{2})$. - $\arg(z_2) = \tan^{-1}(\frac{4}{-2}) = \tan^{-1}(-2)$, but since real part negative and imaginary positive, $\arg(z_2) = \pi - \tan^{-1}(2)$. - $\arg(z_3) = \arg(4i) = \frac{\pi}{2}$. 13. Substitute: $$2 \arg(z_2) - \arg(z_1) + \frac{\pi}{2} = 2\left(\pi - \tan^{-1}(2)\right) - \tan^{-1}\left(\frac{1}{2}\right) + \frac{\pi}{2} = 2\pi - 2 \tan^{-1}(2) - \tan^{-1}\left(\frac{1}{2}\right) + \frac{\pi}{2}.$$ 14. Using $\tan^{-1}(2) + \tan^{-1}(\frac{1}{2}) = \frac{\pi}{2}$, simplify: $$2\pi - 2 \tan^{-1}(2) - \tan^{-1}\left(\frac{1}{2}\right) + \frac{\pi}{2} = 2\pi - \left(2 \tan^{-1}(2) + \tan^{-1}\left(\frac{1}{2}\right)\right) + \frac{\pi}{2} = 2\pi - \frac{3\pi}{2} + \frac{\pi}{2} = \frac{\pi}{2}.$$ 15. Hence, $\arg(z_3) = 2 \arg(z_2) - \arg(z_1) + \frac{\pi}{2}$ is verified. 16. **Problem (d):** Represent $z_1$, $z_2$, and $z_3$ on an Argand diagram. 17. Plot points: - $z_1 = (2,1)$ - $z_2 = (-2,4)$ - $z_3 = (0,4)$ 18. **Problem (e):** Expand $\ln(1 + x)$ to five terms. 19. Using Taylor series: $$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} + \cdots$$ 20. **Problem (f):** Using expansion in (e), find expansion for $\ln\left(\frac{1}{1 - x}\right)$. 21. Note: $$\ln\left(\frac{1}{1 - x}\right) = -\ln(1 - x).$$ 22. Expand $\ln(1 - x)$: $$\ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} + \cdots$$ 23. So: $$\ln\left(\frac{1}{1 - x}\right) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} + \cdots$$ 24. **Problem (g):** Power series for $\ln\left(\frac{1 + x}{1 - x}\right)$. 25. Using properties: $$\ln\left(\frac{1 + x}{1 - x}\right) = \ln(1 + x) - \ln(1 - x).$$ 26. Substitute expansions: $$\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}\right) - \left(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5}\right) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \cdots$$ 27. So the power series is: $$\ln\left(\frac{1 + x}{1 - x}\right) = 2 \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1} = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots\right).$$ 28. **Problem (h):** Evaluate $$\int_0^1 \ln\left(\frac{1 + t}{1 - t}\right) dt.$$ 29. Substitute series: $$\int_0^1 2 \sum_{n=0}^\infty \frac{t^{2n+1}}{2n+1} dt = 2 \sum_{n=0}^\infty \frac{1}{2n+1} \int_0^1 t^{2n+1} dt = 2 \sum_{n=0}^\infty \frac{1}{2n+1} \cdot \frac{1}{2n+2}.$$ 30. Simplify: $$= 2 \sum_{n=0}^\infty \frac{1}{(2n+1)(2n+2)}.$$ 31. Using partial fractions: $$\frac{1}{(2n+1)(2n+2)} = \frac{1}{2n+1} - \frac{1}{2n+2}.$$ 32. The series telescopes: $$2 \sum_{n=0}^\infty \left(\frac{1}{2n+1} - \frac{1}{2n+2}\right) = 2 \left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots \right) = 2 \ln 2.$$ 33. Therefore, $$\int_0^1 \ln\left(\frac{1 + t}{1 - t}\right) dt = 2 \ln 2.$$