1. Problem 18. State the problem: In the Argand diagram the point P represents the complex number $z$ and we are given $|z-1-i|=\sqrt{2}$. 2. Interpretation and locus: The equation $|z-1-i|=\sqrt{2}$ is the circle with centre $1+i$ and radius $\sqrt{2}$. 3. Centre and radius computation: The centre is $1+i$ and the radius is $\sqrt{2}$. 4. Distance from origin to centre: The distance from the origin to the centre is $|1+i|=\sqrt{1^2+1^2}=\sqrt{2}$. 5. Greatest modulus: The maximum value of $|z|$ for $z$ on the circle is the distance from the origin to the centre plus the radius, so $|z|_{\max}=\sqrt{2}+\sqrt{2}=2\sqrt{2}$. 6. Least modulus: The minimum value of $|z|$ is the absolute value of the difference of the distance and the radius, so $|z|_{\min}=|\sqrt{2}-\sqrt{2}|=0$. 7. Points attaining extrema: The point giving $|z|_{\min}=0$ is $z=0$, since $|0-1-i|=\sqrt{2}$. 8. The point giving $|z|_{\max}=2\sqrt{2}$ is $z=2+2i$, since $|2+2i-1-i|=|1+i|=\sqrt{2}$ and $|2+2i|=2\sqrt{2}$. 9. Problem 19. State the problem: Prove the non-real cube roots of unity are $-\tfrac{1}{2}\pm i\tfrac{\sqrt{3}}{2}$ and show that, for $A,B$ those roots and $C$ the point $-2$, the area of the sector of the circle with centre $C$ through $A,B$ bounded by $CA,CB$ and the minor arc $AB$ is $\tfrac{1}{2}\pi$. 10. Cube roots of unity: Solve $z^3=1$ which yields $z= e^{2\pi i k/3}$ for $k=0,1,2$. The non-real roots are $e^{2\pi i/3}$ and $e^{4\pi i/3}$, which equal $-\tfrac{1}{2}\pm i\tfrac{\sqrt{3}}{2}$. 11. Coordinates: Take $A=e^{2\pi i/3}=(-\tfrac{1}{2},\tfrac{\sqrt{3}}{2})$ and $B=e^{4\pi i/3}=(-\tfrac{1}{2},-\tfrac{\sqrt{3}}{2})$ and $C=-2$ which is $(-2,0)$. 12. Radius of circle centre C through A: Compute $|CA|^2=( -\tfrac{1}{2}+2)^2+(\tfrac{\sqrt{3}}{2}-0)^2=(\tfrac{3}{2})^2+(\tfrac{\sqrt{3}}{2})^2=\tfrac{9}{4}+\tfrac{3}{4}=3$, so the radius is $\sqrt{3}$. 13. Central angle: Compute the dot product $CA\cdot CB=(\tfrac{3}{2})^2-(\tfrac{\sqrt{3}}{2})^2=\tfrac{9}{4}-\tfrac{3}{4}=\tfrac{3}{2}$ and $|CA|=|CB|=\sqrt{3}$, so $\cos\angle ACB=\dfrac{\tfrac{3}{2}}{(\sqrt{3})(\sqrt{3})}=\tfrac{1}{2}$, hence $\angle ACB=\pi/3$. 14. Sector area: The area of the sector is $\tfrac{1}{2}r^2\theta=\tfrac{1}{2}\cdot(\sqrt{3})^2\cdot\tfrac{\pi}{3}=\tfrac{1}{2}\pi$. 15. Problem 20. State the problem: Find the roots of $z^4+4=0$ and express $z^4+4$ as a product of two quadratics with real coefficients. 16. Solve equation: Write $z^4=-4=4e^{i(\pi+2k\pi)}$, so $z=4^{1/4}e^{i(\pi+2k\pi)/4}=\sqrt{2}\,e^{i(\pi/4+k\pi/2)}$ for $k=0,1,2,3$. 17. List roots explicitly: The four roots are $\sqrt{2}e^{i\pi/4}=1+i$, $\sqrt{2}e^{i3\pi/4}=-1+i$, $\sqrt{2}e^{i5\pi/4}=-1-i$, and $\sqrt{2}e^{i7\pi/4}=1-i$. 18. Factorisation into quadratics: Pair conjugates to get $(z-(1+i))(z-(1-i))=z^2-2z+2$ and $(z-(-1+i))(z-(-1-i))=z^2+2z+2$, hence $$z^4+4=(z^2-2z+2)(z^2+2z+2).$$ 19. Problem 21. State the problem: State de Moivre's theorem and hence express $\cos5\theta$ as a polynomial in $\cos\theta$, and prove the given factorisation by considering roots of $\cos5\theta=0$. 20. De Moivre's theorem: For integer $n$ we have $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. 21. Compute $\cos5\theta$: Using de Moivre and binomial expansion, taking the real part gives $$\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta.$$ 22. Roots of $\cos5\theta=0$: Solve $\cos5\theta=0$ gives $5\theta=\tfrac{\pi}{2}+k\pi$, so $\theta=\tfrac{\pi}{10}+\tfrac{k\pi}{5}$ and the distinct values of $\cos\theta$ that are zeros are $0,\pm\cos\tfrac{\pi}{10},\pm\cos\tfrac{3\pi}{10}$. 23. Factorisation as polynomial in $\cos\theta$: A degree 5 polynomial with leading coefficient 16 and those five zeros factors as $$\cos5\theta=16\cos\theta\bigl(\cos\theta-\cos\tfrac{\pi}{10}\bigr)\bigl(\cos\theta-\cos\tfrac{3\pi}{10}\bigr)\bigl(\cos\theta-\cos\tfrac{7\pi}{10}\bigr)\bigl(\cos\theta-\cos\tfrac{9\pi}{10}\bigr),$$ and using $\cos\tfrac{7\pi}{10}=-\cos\tfrac{3\pi}{10}$ and $\cos\tfrac{9\pi}{10}=-\cos\tfrac{\pi}{10}$ this is the stated factorisation. 24. Problem 22. State the problem: Prove de Moivre's theorem for positive integers, deduce $z^n-z^{-n}=2i\sin n\theta$ and prove $16\sin^5\theta=\sin5\theta-5\sin3\theta+10\sin\theta$, then solve $4\sin^5\theta+\sin5\theta=0$ on $[0,2\pi]$. 25. De Moivre proof outline: For positive integer $n$ the result follows by induction using multiplication formulas for sines and cosines, giving $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. 26. Negative integers and subtraction: Assuming the identity holds for negative integers as well, set $z=\cos\theta+i\sin\theta$ and compute $z^n-z^{-n}=(\cos n\theta+i\sin n\theta)-(\cos n\theta-i\sin n\theta)=2i\sin n\theta$. 27. Fifth-power sine identity: From de Moivre expand the imaginary part of $(\cos\theta+i\sin\theta)^5$ to get $$\sin5\theta=16\sin^5\theta-20\sin^3\theta+5\sin\theta.$$ Replace $\sin^3\theta$ by $\tfrac{3\sin\theta-\sin3\theta}{4}$ or rearrange to obtain $$16\sin^5\theta=\sin5\theta-5\sin3\theta+10\sin\theta.$$ 28. Solve the equation $4\sin^5\theta+\sin5\theta=0$: Substitute the identity to get $4\sin^5\theta+\sin5\theta=5\sin5\theta-5\sin3\theta+10\sin\theta$ which simplifies (directly via algebra) to $\sin\theta\bigl(4\sin^4\theta-4\sin^2\theta+1\bigr)=0$. 29. Reduce quartic factor: Let $u=\sin^2\theta$, then $4u^2-4u+1=(2u-1)^2$, so $u=\tfrac{1}{2}$ is a double root, giving $\sin\theta=0$ or $\sin\theta=\pm\tfrac{1}{\sqrt{2}}$. 30. Solutions in $[0,2\pi]$: Thus $\sin\theta=0$ gives $\theta=0,\pi,2\pi$ and $\sin\theta=1/\sqrt{2}$ gives $\theta=\tfrac{\pi}{4},\tfrac{3\pi}{4}$ and $\sin\theta=-1/\sqrt{2}$ gives $\theta=\tfrac{5\pi}{4},\tfrac{7\pi}{4}$. 31. Final answer summary: Problem 18 locus is the circle centre $1+i$ radius $\sqrt{2}$ with $|z|_{\min}=0$ and $|z|_{\max}=2\sqrt{2}$. 32. Final answer summary: Problem 19 non-real cube roots are $-\tfrac{1}{2}\pm i\tfrac{\sqrt{3}}{2}$ and the specified sector area is $\tfrac{1}{2}\pi$. 33. Final answer summary: Problem 20 roots are $1\pm i$ and $-1\pm i$ and $z^4+4=(z^2-2z+2)(z^2+2z+2)$. 34. Final answer summary: Problem 21 de Moivre and $\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta$ and the stated factorisation follows from the zeros. 35. Final answer summary: Problem 22 de Moivre gives $z^n-z^{-n}=2i\sin n\theta$, the quintic sine identity holds and the solutions of $4\sin^5\theta+\sin5\theta=0$ on $[0,2\pi]$ are $0,\tfrac{\pi}{4},\tfrac{3\pi}{4},\pi,\tfrac{5\pi}{4},\tfrac{7\pi}{4},2\pi$.