Problem statement: Work through the listed complex-number exercises: simplify complex fractions (a)–(j), simplify a general fraction in 5, solve quadratics in 6, find complex square roots in 7, form quadratics from conjugate roots in 8, prove the relations in 9, solve the cubics and quartics in 10–11, determine coefficients in 12–13, and answer loci and roots in later questions as indicated. 1. Simplify (a)–(j). 1. (a) Simplify $\frac{20}{3+i}$. Multiply numerator and denominator by the conjugate $3 - i$. Compute $\frac{20}{3+i}=\frac{20(3 - i)}{(3+i)(3 - i)}$. Compute numerator $20(3 - i)=60 - 20i$. Denominator $(3+i)(3 - i)=3^2 +1^2=10$. Hence $\frac{20}{3+i}=\frac{60 - 20i}{10}=6 - 2i$. 1. (b) Simplify $\frac{4}{1+i}$. Multiply by $1 - i$. $\frac{4}{1+i}=\frac{4(1 - i)}{(1+i)(1 - i)}=\frac{4 - 4i}{2}=2 - 2i$. 1. (c) Simplify $\frac{21}{1-2i}$. Multiply by $1 + 2i$. $\frac{21}{1-2i}=\frac{21(1+2i)}{1+4}=\frac{21 + 42i}{5}=\frac{21}{5}+\frac{42}{5}i$. 1. (d) Simplify $\frac{1}{3-2i}$. Multiply by $3 + 2i$. $\frac{1}{3-2i}=\frac{3+2i}{9+4}=\frac{3+2i}{13}$. 1. (e) Simplify $\frac{5i}{1+2i}$. Multiply by $1 - 2i$. $\frac{5i}{1+2i}=\frac{5i(1 - 2i)}{1+4}=\frac{5i -10i^2}{5}=\frac{10+5i}{5}=2 + i$. 1. (f) Simplify $\frac{5}{4-3i}$. Multiply by $4 + 3i$. $\frac{5}{4-3i}=\frac{5(4+3i)}{16+9}=\frac{20+15i}{25}=\frac{4}{5}+\frac{3}{5}i$. 1. (g) Simplify $\frac{2+3i}{1-i}$. Multiply by $1 + i$. $\frac{2+3i}{1-i}=\frac{(2+3i)(1+i)}{2}=\frac{-1+5i}{2}=-\frac{1}{2}+\frac{5}{2}i$. 1. (h) Simplify $\frac{3+2i}{1+2i}$. Multiply by $1 - 2i$. $\frac{3+2i}{1+2i}=\frac{(3+2i)(1 - 2i)}{5}=\frac{7 -4i}{5}=\frac{7}{5}-\frac{4}{5}i$. 1. (i) Simplify $\frac{3+2i}{3-2i}$. Multiply by $3 + 2i$. $\frac{3+2i}{3-2i}=\frac{(3+2i)(3+2i)}{13}=\frac{5+12i}{13}=\frac{5}{13}+\frac{12}{13}i$. 1. (j) Simplify $\frac{4i-3}{2+3i}$. Multiply by $2 - 3i$. $\frac{4i-3}{2+3i}=\frac{(4i-3)(2-3i)}{13}=\frac{6+17i}{13}=\frac{6}{13}+\frac{17}{13}i$. 2. Problem 5: Simplify $\frac{a+ib}{b - ai}$. Multiply numerator and denominator by the conjugate $b + ai$. $\frac{a+ib}{b - ai}=\frac{(a+ib)(b+ai)}{b^2 + a^2}$. Expand numerator: $(a+ib)(b+ai)=ab + a^2 i + ib^2 + i^2 ab$. Since $i^2=-1$ the $ab$ terms cancel and numerator becomes $i(a^2 + b^2)$. So $\frac{a+ib}{b - ai}=\frac{i(a^2+b^2)}{a^2+b^2}=i$. 3. Problem 6: Solve the quadratics (assume equal to 0 where missing). 3. (a) $x^2 + 25 = 0$. $x^2=-25$ so $x=\pm 5i$. 3. (b) $2x^2 + 32 = 0$. $2x^2=-32$ so $x^2=-16$ and $x=\pm 4i$. 3. (c) $4x^2 + 9 = 0$. $4x^2=-9$ so $x=\pm \frac{3}{2}i$. 3. (d) $x^2 + 2x + 5 = 0$. Use quadratic formula $x=\dfrac{-2\pm\sqrt{4-20}}{2}=\dfrac{-2\pm\sqrt{-16}}{2}=-1\pm 2i$. 3. (e) $x^2 - 4x + 5 = 0$. $x=\dfrac{4\pm\sqrt{16-20}}{2}=\dfrac{4\pm\sqrt{-4}}{2}=2\pm i$. 3. (f) $2x^2 + x + 1 = 0$. $x=\dfrac{-1\pm\sqrt{1-8}}{4}=\dfrac{-1\pm i\sqrt{7}}{4}$. 4. Problem 7: Find square roots. 4. (a) Square roots of $5+12i$. Compute modulus $r=\sqrt{5^2+12^2}=13$. Real part $u=\sqrt{\frac{r+5}{2}}=\sqrt{\frac{13+5}{2}}=\sqrt{9}=3$. Imag part $v=\operatorname{sgn}(12)\sqrt{\frac{r-5}{2}}=\sqrt{\frac{13-5}{2}}=\sqrt{4}=2$. Roots: $3+2i$ and $-3-2i$. 4. (b) Square roots of $15+8i$. $r=\sqrt{15^2+8^2}=17$. $u=\sqrt{\frac{17+15}{2}}=\sqrt{16}=4$. $v=\sqrt{\frac{17-15}{2}}=\sqrt{1}=1$. Roots: $4+i$ and $-4-i$. 4. (c) Square roots of $7 -24i$. $r=\sqrt{7^2+(-24)^2}=25$. $u=\sqrt{\frac{25+7}{2}}=\sqrt{16}=4$. $v=\operatorname{sgn}(-24)\sqrt{\frac{25-7}{2}}=-\sqrt{9}=-3$. Roots: $4-3i$ and $-4+3i$. 5. Problem 8: Quadratics with given roots (use $x^2-(r_1+r_2)x + r_1r_2$). 5. (a) Roots $3i, -3i$ give sum $0$ and product $9$. Quadratic $x^2+9=0$. 5. (b) Roots $1+2i,1-2i$ give sum $2$ and product $5$. Quadratic $x^2-2x+5=0$. 5. (c) Roots $2+i,2-i$ give sum $4$ and product $5$. Quadratic $x^2-4x+5=0$. 5. (d) Roots $2+3i,2-3i$ give sum $4$ and product $13$. Quadratic $x^2-4x+13=0$. 5. (e) Roots $3+4i,3-4i$ give sum $6$ and product $25$. Quadratic $x^2-6x+25=0$. 5. (f) Roots $3+5i,3-5i$ give sum $6$ and product $34$. Quadratic $x^2-6x+34=0$. 6. Problem 9: If $a+ib$ is a root of $x^2 + cx + d = 0$ with real $c,d$, show $a^2+b^2=d$ and $2a+c=0$. Complex coefficients are real so conjugate $a-ib$ is also a root. Sum of roots = $-c$ so $(a+ib)+(a-ib)=2a=-c$ which gives $2a+c=0$. Product of roots = $d$ so $(a+ib)(a-ib)=a^2+b^2=d$. Thus the required relations hold. 7. Problem 10: Solve the polynomials. 7. (a) $x^3 -7x^2 +19x -13 = 0$. Test $x=1$: $1-7+19-13=0$ so $x=1$ is a root. Divide out $(x-1)$ to get $x^2-6x+13=0$. Solve $x=\dfrac{6\pm\sqrt{36-52}}{2}=3\pm 2i$. Roots: $1,3+2i,3-2i$. 7. (b) $2x^3 -2x^2 -3x -2 = 0$. Test $x=2$: gives $0$ so $x=2$ is root. Divide to get $2x^2+2x+1=0$. $x=\dfrac{-2\pm\sqrt{4-8}}{4}=\dfrac{-1\pm i}{2}$. Roots: $2,\dfrac{-1+i}{2},\dfrac{-1-i}{2}$. 7. (c) $x^3 +3x^2 +5x +3 = 0$. Test $x=-1$: gives $0$ so $x=-1$ is root. Divide to get $x^2+2x+3=0$. $x=-1\pm i\sqrt{2}$. Roots: $-1,-1+i\sqrt{2},-1-i\sqrt{2}$. 7. (d) $4x^4 -20x^3 +37x^2 -31x +10 = 0$. Test $x=1$: gives $0$ so $x=1$ is root. Divide to get $4x^3-16x^2+21x-10$. Test $x=2$: gives $0$ so $x=2$ is root. Divide to get $4x^2-8x+5=0$. Solve $x=\dfrac{8\pm\sqrt{64-80}}{8}=1\pm \tfrac{i}{2}$. Roots: $1,2,1+\tfrac{i}{2},1-\tfrac{i}{2}$. 7. (e) $5x^4 +8x^3 -8x -5 = 0$. Test $x=1$: gives $0$ so $x=1$ is root. Divide to get $5x^3+13x^2+13x+5$. Test $x=-1$: gives $0$ so $x=-1$ is root. Divide to get $5x^2+8x+5=0$. Solve $x=\dfrac{-8\pm\sqrt{64-100}}{10}= -\dfrac{4}{5}\pm \dfrac{3}{5}i$. Roots: $1,-1,-\dfrac{4}{5}+\dfrac{3}{5}i,-\dfrac{4}{5}-\dfrac{3}{5}i$. 8. Problem 11: Factorization and roots. Given $5x^4 -14x^3 +18x^2 +40x +16=(x^2 -4x +8)(ax^2+bx+c)$. Multiply RHS and equate coefficients. Coefficient of $x^4$: $a=5$. x^3: $b-4a=-14$ so $b=6$. x^2: $c-4b+8a=18$ so $c-24+40=18$ giving $c=2$. Check $x$ coefficient $-4c+8b=-8+48=40$ and constant $8c=16$ agree. So $a=5,b=6,c=2$. Solve factors $x^2-4x+8=0$ gives $x=2\pm 2i$. Solve $5x^2+6x+2=0$ gives $x= -\dfrac{3}{5}\pm \dfrac{1}{5}i$. 9. Problem 12: If $3-2i$ and $1+i$ are roots of $ax^4+bx^3+cx^2+dx+e=0$ with real coefficients, their conjugates $3+2i$ and $1-i$ are also roots. Form polynomial as $(x-(3-2i))(x-(3+2i))(x-(1+i))(x-(1-i))$. Pairwise multiply: $(x^2-6x+13)(x^2-2x+2)$. Expand to get $x^4-8x^3+27x^2-38x+26=0$. Thus $a=1,b=-8,c=27,d=-38,e=26$. 10. Problem 13: Factor $x^3-1=(x-1)(ax^2+bx+c)$ and find roots of $x^3=1$. Divide to find $a=1,b=1,c=1$ so $x^3-1=(x-1)(x^2+x+1)$. Solve $x^2+x+1=0$ gives $x= -\tfrac{1}{2}\pm i\tfrac{\sqrt{3}}{2}$. So cube roots of unity are $1, -\tfrac{1}{2}+i\tfrac{\sqrt{3}}{2}, -\tfrac{1}{2}-i\tfrac{\sqrt{3}}{2}$. 11. Problem 18: Locus $|z-1-i|=\sqrt{2}$. This is a circle centered at $1+i$ with radius $\sqrt{2}$. Distance from origin to center is $|1+i|=\sqrt{2}$. Maximum $|z|$ on the circle is center distance plus radius $\sqrt{2}+\sqrt{2}=2\sqrt{2}$. Minimum $|z|$ is center distance minus radius $\sqrt{2}-\sqrt{2}=0$. Hence greatest value $2\sqrt{2}$ and least value $0$. 12. Problem 20: Solve $z^4+4=0$ and factor into quadratics with real coefficients. Write $z^4=-4=4e^{i\pi}$. Fourth roots have modulus $4^{1/4}=\sqrt{2}$ and arguments $\dfrac{\pi+2k\pi}{4}=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$ for $k=0,1,2,3$. Thus roots are $\sqrt{2}e^{i\pi/4}=1+i$, $\sqrt{2}e^{i3\pi/4}=-1+i$, $\sqrt{2}e^{i5\pi/4}=-1-i$, $\sqrt{2}e^{i7\pi/4}=1-i$. Group conjugate pairs to form quadratics: $z^4+4=(z^2+2z+2)(z^2-2z+2)$. Final answers summary (boxed): - (1a) $6-2i$. - (1b) $2-2i$. - (1c) $\frac{21}{5}+\frac{42}{5}i$. - (1d) $\frac{3+2i}{13}$. - (1e) $2+i$. - (1f) $\frac{4}{5}+\frac{3}{5}i$. - (1g) $-\tfrac{1}{2}+\tfrac{5}{2}i$. - (1h) $\tfrac{7}{5}-\tfrac{4}{5}i$. - (1i) $\tfrac{5}{13}+\tfrac{12}{13}i$. - (1j) $\tfrac{6}{13}+\tfrac{17}{13}i$. - (5) $i$. - (6a) $\pm 5i$, (6b) $\pm 4i$, (6c) $\pm \tfrac{3}{2}i$, (6d) $-1\pm 2i$, (6e) $2\pm i$, (6f) $\dfrac{-1\pm i\sqrt{7}}{4}$. - (7a) $3+2i,\ -3-2i$. - (7b) $4+i,\ -4-i$. - (7c) $4-3i,\ -4+3i$. - (8a) $x^2+9=0$. - (8b) $x^2-2x+5=0$. - (8c) $x^2-4x+5=0$. - (8d) $x^2-4x+13=0$. - (8e) $x^2-6x+25=0$. - (8f) $x^2-6x+34=0$. - (9) $a^2+b^2=d$ and $2a+c=0$. - (10a) $1,3\pm 2i$. - (10b) $2,\dfrac{-1\pm i}{2}$. - (10c) $-1,-1\pm i\sqrt{2}$. - (10d) $1,2,1\pm \tfrac{i}{2}$. - (10e) $1,-1,-\dfrac{4}{5}\pm \dfrac{3}{5}i$. - (11) $a=5,b=6,c=2$ and roots $2\pm 2i, -\tfrac{3}{5}\pm \tfrac{1}{5}i$. - (12) $a=1,b=-8,c=27,d=-38,e=26$. - (13) $a=b=c=1$ for the quadratic factor and cube roots $1, -\tfrac{1}{2}\pm i\tfrac{\sqrt{3}}{2}$. - (18) locus: circle centre $1+i$, radius $\sqrt{2}$, min $|z|=0$, max $|z|=2\sqrt{2}$. - (20) roots $1\pm i,-1\pm i$ and factorization $(z^2+2z+2)(z^2-2z+2)$. (Worked details are shown step by step above for each numbered item.)