Complex Numbers

Q1 (10)

(a) Numerator: $1+i\sqrt3$ has modulus $\sqrt{1+3}=2$, argument $\arctan(\sqrt3/1)=\pi/3$. So $1+i\sqrt3 = 2e^{i\pi/3}$. Then $(1+i\sqrt3)^5 = 2^5 e^{i5\pi/3} = 32e^{i5\pi/3}$. (2)

Denominator: $1-i$ has modulus $\sqrt2$, argument $-\pi/4$. So $1-i=\sqrt2\,e^{-i\pi/4}$. Then $(1-i)^3 = 2^{3/2}e^{-i3\pi/4} = 2\sqrt2\,e^{-i3\pi/4}$. (2)

(b) $z = \frac{32e^{i5\pi/3}}{2\sqrt2\,e^{-i3\pi/4}} = \frac{32}{2\sqrt2}\,e^{i(5\pi/3 + 3\pi/4)}.$ Modulus $= 16/\sqrt2 = 8\sqrt2$. (2) Angle $= 5\pi/3 + 3\pi/4 = \dfrac{20\pi+9\pi}{12} = \dfrac{29\pi}{12}$. Reduce mod $2\pi$: $29\pi/12 - 24\pi/12 = 5\pi/12 \in(-\pi,\pi]$. (2) $z = 8\sqrt2\,e^{i5\pi/12}.$

(c) $\text{Re}(z)=8\sqrt2\cos(5\pi/12)$, $\text{Im}(z)=8\sqrt2\sin(5\pi/12)$. Using $\cos75^\circ=\frac{\sqrt6-\sqrt2}{4}$, $\sin75^\circ=\frac{\sqrt6+\sqrt2}{4}$: $\text{Re}(z)=8\sqrt2\cdot\frac{\sqrt6-\sqrt2}{4}=2\sqrt2(\sqrt6-\sqrt2)=4\sqrt3-4,$ $\text{Im}(z)=2\sqrt2(\sqrt6+\sqrt2)=4\sqrt3+4.$ (2)

Q2 (10)

(a) Cube roots differ by $2\pi/3$ in argument, all with modulus $2$: $z_2 = 2e^{i(\pi/9+2\pi/3)} = 2e^{i7\pi/9}$, $z_3 = 2e^{i(\pi/9+4\pi/3)} = 2e^{i13\pi/9}$ (or $2e^{-i5\pi/9}$). (3)

(b) $w = z_1^3 = (2e^{i\pi/9})^3 = 8e^{i\pi/3} = 8(\cos60^\circ + i\sin60^\circ) = 4 + 4\sqrt3\,i.$ (4)

(c) All three roots have modulus $2$ and arguments equally spaced by $120^\circ$, so they lie on a circle radius $2$ at vertices of an equilateral triangle (rotational symmetry). (1) Side length $s = 2R\sin(60^\circ) = 2\cdot2\cdot\frac{\sqrt3}{2} = 2\sqrt3$. Area $=\frac{\sqrt3}{4}s^2 = \frac{\sqrt3}{4}\cdot12 = 3\sqrt3$. (2)

Q3 (10)

(a) By De Moivre, $\cos5\theta + i\sin5\theta = (\cos\theta+i\sin\theta)^5$. Expand (binomial), take real part with $c=\cos\theta,s=\sin\theta$: $\cos5\theta = c^5 - 10c^3s^2 + 5cs^4.$ (3) Substitute $s^2 = 1-c^2$: $= c^5 - 10c^3(1-c^2) + 5c(1-c^2)^2 = c^5 -10c^3+10c^5 +5c(1-2c^2+c^4).$ $= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5 = 16c^5 - 20c^3 + 5c.\ \blacksquare$ (2)

(b) $16x^5-20x^3+5x = 0$ with $x=\cos\theta$ means $\cos5\theta=0$, i.e. $5\theta = 90^\circ + 180^\circ k$, so $\theta = 18^\circ,54^\circ,90^\circ,126^\circ,162^\circ$ (giving distinct $x$). (2) Roots: $x = \cos18^\circ,\cos54^\circ, 0, \cos126^\circ=-\cos54^\circ, \cos162^\circ=-\cos18^\circ$. So $x\in\{0,\pm\cos18^\circ,\pm\cos54^\circ\}$. (1) Factor out $x$: $16x^4-20x^2+5=0$, $x^2 = \frac{20\pm\sqrt{400-320}}{32}=\frac{20\pm\sqrt{80}}{32}=\frac{5\pm\sqrt5}{8}$. $\cos36^\circ = \sin54^\circ$; note $\cos54^\circ=\sin36^\circ$. Take largest positive root $\cos18^\circ$... but we want $\cos36^\circ$: use $\cos36^\circ$ satisfies $\cos36^\circ = 1-2\sin^2 18^\circ$; alternatively directly, $\cos^2 36^\circ = \frac{3+\sqrt5}{8}$ gives $\cos36^\circ = \frac{\sqrt5+1}{4}$. (2)

Q4 (10)

(a) $P$ has real coefficients, so non-real roots occur in conjugate pairs; hence $\overline{1+2i}=1-2i$ is a root. (2)

(b) $(z-(1+2i))(z-(1-2i)) = z^2 - 2z + (1+4) = z^2 - 2z + 5.$ (3)

(c) Divide: $P(z) = (z^2-2z+5)(z^2 + az + b)$. Expand and match: $z^2$ coeff: $5 + b - 2a = 14$; $z$ coeff: $5a - 2b = -20$; constant $5b=25\Rightarrow b=5$. Then $5+5-2a=14\Rightarrow a=-2$. Check $z$: $5(-2)-2(5)=-20$ ✓. (3) So other factor $z^2 - 2z + 5$ again — identical. Roots of $z^2-2z+5=0$: $z=1\pm2i$. Thus $P(z)=(z^2-2z+5)^2$, all four roots: $1+2i,\ 1-2i$ each double. (2)

Q5 (10)

(a) These are the 7 roots of $z^7=1$, i.e. roots of $z^7-1=(z-1)(z^6+\cdots+1)$. Sum of all 7 roots $=0$ (coeff of $z^6$ in $z^7-1$ is $0$). So $1+\omega+\cdots+\omega^6 = 0$. (3)

(b) $S+T = \omega+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6 = -1$ from (a). (2) $ST = (\omega+\omega^2+\omega^4)(\omega^3+\omega^5+\omega^6)$. Expand (exponents mod 7):

• $\omega\cdot\omega^3=\omega^4$, $\omega\cdot\omega^5=\omega^6$, $\omega\cdot\omega^6=\omega^7=1$
• $\omega^2\cdot\omega^3=\omega^5$, $\omega^2\cdot\omega^5=\omega^7=1$, $\omega^2\cdot\omega^6=\omega^8=\omega$
• $\omega^4\cdot\omega^3=\omega^7=1$, $\omega^4\cdot\omega^5=\omega^9=\omega^2$, $\omega^4\cdot\omega^6=\omega^{10}=\omega^3$

Sum $= 3 + (\omega+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6) = 3 + (-1) = 2.$ (3)

(c) $S,T$ are roots of $u^2 - (S+T)u + ST = u^2 + u + 2 = 0$: $u = \frac{-1\pm\sqrt{1-8}}{2} = \frac{-1\pm i\sqrt7}{2}$. Since $\text{Im}(S)=\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}>0$, take $S = \frac{-1 + i\sqrt7}{2}.$ (2)

[
  {"claim":"Q1 z equals 8*sqrt(2)*exp(i*5pi/12) giving Re=4sqrt3-4, Im=4sqrt3+4","code":"z=(1+I*sqrt(3))**5/(1-I)**3; z=simplify(z); result = simplify(re(z)-(4*sqrt(3)-4))==0 and simplify(im(z)-(4*sqrt(3)+4))==0"},
  {"claim":"Q2 w = z1^3 = 4+4sqrt3 i","code":"z1=2*exp(I*pi/9); w=expand_complex(z1**3); result = simplify(w-(4+4*sqrt(3)*I))==0"},
  {"claim":"Q3 identity cos5θ=16c^5-20c^3+5c","code":"th=symbols('th'); lhs=cos(5*th); rhs=16*cos(th)**5-20*cos(th)**3+5*cos(th); result = simplify(expand_trig(lhs)-rhs)==0"},
  {"claim":"Q3 cos36 = (sqrt5+1)/4","code":"result = simplify(cos(pi/5)-(sqrt(5)+1)/4)==0"},
  {"claim":"Q4 P(z)=(z^2-2z+5)^2","code":"z=symbols('z'); P=z**4-4*z**3+14*z**2-20*z+25; result = simplify(P-(z**2-2*z+5)**2)==0"},
  {"claim":"Q5 S=(-1+i sqrt7)/2 with omega=exp(2pi i/7)","code":"w=exp(2*pi*I/7); S=w+w**2+w**4; result = simplify(S-(-1+I*sqrt(7))/2)==0"}
]