Complex Numbers

Question 1 (10)

(a) Series (1 mark for all three): $e^x=\sum_{k=0}^\infty \frac{x^k}{k!},\quad \cos x=\sum \frac{(-1)^m x^{2m}}{(2m)!},\quad \sin x=\sum \frac{(-1)^m x^{2m+1}}{(2m+1)!}$ Substitute $x=i\theta$ (1): $e^{i\theta}=\sum_{k=0}^\infty \frac{(i\theta)^k}{k!}=1+i\theta+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\cdots$ Use powers of $i$: $i^0=1,i^1=i,i^2=-1,i^3=-i,i^4=1$ (1). Then: $=\Big(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\Big)+i\Big(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots\Big)$ (separating even/odd $k$: 2 marks). Recognise real part $=\cos\theta$, imaginary part $=\sin\theta$ (1): $e^{i\theta}=\cos\theta+i\sin\theta \quad(1)$

(b) $e^{i\pi}=\cos\pi+i\sin\pi=-1$ (1), so $e^{i\pi}+1=0$ (1). Geometrically: rotating the point $1$ by $\pi$ radians about the origin lands at $-1$ on the negative real axis (1).

Question 2 (12)

(a) Statement (1): $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. Base $n=1$ trivially true (1). Assume true for $n=k$ (1): $(\cos\theta+i\sin\theta)^k=\cos k\theta+i\sin k\theta$. Multiply both sides by $(\cos\theta+i\sin\theta)$ (1): $=(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)$ Expand and use angle-addition formulas (1): $=\cos(k\theta+\theta)+i\sin(k\theta+\theta)=\cos(k+1)\theta+i\sin(k+1)\theta$ Hence true for $k+1$; by induction true for all positive integers $n$ (1).

(b) $(\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta$. Expand LHS (1): $\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta$ Real part (1): $\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta$. Sub $\sin^2\theta=1-\cos^2\theta$ (1): $\cos3\theta=4\cos^3\theta-3\cos\theta \quad(1)$

(c) For fractional $n$ the expression $(\cos\theta+i\sin\theta)^{p/q}$ is multivalued: adding $2\pi k$ to $\theta$ leaves the base unchanged but changes the fractional-power result, so De Moivre's single-value form captures only one of the $q$ distinct roots (2).

Question 3 (12)

(a) $z=-8i$: modulus $8$, argument $-\pi/2$ (or $3\pi/2$) (1). Write $z=8\,e^{i(-\pi/2+2\pi k)}$ (1). Cube roots: modulus $2$, arguments $\theta_k=\dfrac{-\pi/2+2\pi k}{3}$ for $k=0,1,2$ (1):

• $k=0$: $2e^{-i\pi/6}=2(\cos(-\pi/6)+i\sin(-\pi/6))=\sqrt3-i$
• $k=1$: $2e^{i\pi/2}=2i$
• $k=2$: $2e^{i7\pi/6}=-\sqrt3-i$

(exponential forms 1.5, rectangular 1.5).

(b) The $n$th roots of unity are $\omega_k=e^{2\pi i k/n}$, $k=0,\dots,n-1$ (1). These are a geometric series with ratio $\omega=e^{2\pi i/n}\ne1$ (1): $\sum_{k=0}^{n-1}\omega^k=\frac{\omega^n-1}{\omega-1}=\frac{1-1}{\omega-1}=0$ since $\omega^n=1$ (2).

(c) They lie equally spaced on the unit circle, forming vertices of a regular $n$-gon, one vertex at $1$ (2).

Question 4 (10)

(a) Multiply by conjugate (1): $\frac{3+2i}{1-4i}\cdot\frac{1+4i}{1+4i}=\frac{(3+2i)(1+4i)}{1+16}$ Numerator: $3+12i+2i+8i^2=3+14i-8=-5+14i$ (1). Denominator $17$ (1). $=-\frac{5}{17}+\frac{14}{17}i \quad(1)$

(b) Let $z_1=a_1+b_1i$, $z_2=a_2+b_2i$. Product (1): $z_1z_2=(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i$ $\overline{z_1z_2}=(a_1a_2-b_1b_2)-(a_1b_2+a_2b_1)i \quad(1)$ Also $\overline{z_1}\,\overline{z_2}=(a_1-b_1i)(a_2-b_2i)=(a_1a_2-b_1b_2)-(a_1b_2+a_2b_1)i$ (1). Equal, QED (1).

(c) $2025=4\cdot506+1$, so $i^{2025}=i$ (1). $i^{-2025}=1/i=-i$, so sum $=i-i=0$ (1).

Question 5 (8)

import cmath, math
def nth_roots(r, theta, n):
    roots = []
    R = r ** (1.0/n)          # modulus of each root = r^(1/n)
    for k in range(n):
        ang = (theta + 2*math.pi*k) / n   # arguments spaced by 2π/n
        roots.append((R, ang))
    return roots

Marks: correct signature/loop (2), modulus formula $R=r^{1/n}$ annotated (2), argument formula $(\theta+2\pi k)/n$ annotated (2), returns $n$ pairs for $k=0..n-1$ (2).

Question 6 (8)

(a) Try rational roots. $p(2)=8-8+18-18=0$ (1), so $x=2$ is the real root (1). Divide: $p(x)=(x-2)(x^2+9) \quad(1)$ $x^2+9=0\Rightarrow x=\pm3i$ (1). Full factorisation: $p(x)=(x-2)(x-3i)(x+3i) \quad(1)$

(b) If $p$ has real coefficients and $p(\alpha)=0$, then taking conjugates $\overline{p(\alpha)}=p(\overline\alpha)=0$ (since conjugation distributes over sums/products and fixes real coefficients). Hence $\overline\alpha$ is also a root — non-real roots pair up (3).

[
  {"claim":"e^{iπ}+1=0","code":"import sympy as sp; result = sp.simplify(sp.exp(sp.I*sp.pi)+1)==0"},
  {"claim":"cos3θ=4cos^3θ-3cosθ","code":"th=sp.symbols('th'); result = sp.simplify(sp.cos(3*th)-(4*sp.cos(th)**3-3*sp.cos(th)))==0"},
  {"claim":"(3+2i)/(1-4i)=-5/17+14/17 i","code":"z=(3+2*sp.I)/(1-4*sp.I); result = sp.simplify(z-(sp.Rational(-5,17)+sp.Rational(14,17)*sp.I))==0"},
  {"claim":"cube roots of -8i are sqrt3-i, 2i, -sqrt3-i","code":"vals=[sp.simplify(v**3) for v in [sp.sqrt(3)-sp.I, 2*sp.I, -sp.sqrt(3)-sp.I]]; result = all(sp.simplify(v-(-8*sp.I))==0 for v in vals)"},
  {"claim":"p(x)=(x-2)(x^2+9) and i^2025+i^-2025=0","code":"x=sp.symbols('x'); p=x**3-2*x**2+9*x-18; f=sp.expand((x-2)*(x**2+9)); c1=sp.simplify(p-f)==0; c2=sp.simplify(sp.I**2025+sp.I**(-2025))==0; result = c1 and c2"}
]