Complex Numbers

Question 1

(a) (5 marks) Series: $e^{x}=\sum_{k\ge0}\frac{x^k}{k!}$. Substitute $x=i\theta$ (valid since the exponential series converges absolutely for all complex arguments — convergence assumption, 1 mark): $e^{i\theta}=\sum_{k\ge0}\frac{(i\theta)^k}{k!}.$ Powers of $i$ cycle $1,i,-1,-i$ (1 mark). Split into even $k=2m$ and odd $k=2m+1$ (1 mark): $=\sum_m\frac{(-1)^m\theta^{2m}}{(2m)!}+i\sum_m\frac{(-1)^m\theta^{2m+1}}{(2m+1)!}=\cos\theta+i\sin\theta.$ Recognition of the two series as $\cos\theta,\sin\theta$ (2 marks).

(b) (4 marks) Statement: $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$ (1). Base $n=1$ trivial (1). Inductive step: assume true for $n$; then $(\cos\theta+i\sin\theta)^{n+1}=(\cos n\theta+i\sin n\theta)(\cos\theta+i\sin\theta)$; expand and use angle-addition to get $\cos(n+1)\theta+i\sin(n+1)\theta$ (2).

(c) (5 marks) $\cos5\theta+i\sin5\theta=(\cos\theta+i\sin\theta)^5$. Expand binomially; take real part (1). Let $c=\cos\theta,s=\sin\theta$: Real part $= c^5 -10c^3s^2+5cs^4$ (2). Substitute $s^2=1-c^2$ (1): $\cos5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta.$ (final, 1)

(d) (6 marks) Denominator $D=(\omega_0^2-\omega^2)+i\gamma\omega$. So $|Z|=\frac{1}{\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}},\qquad \arg(Z)=-\arg(D)=-\arctan\!\frac{\gamma\omega}{\omega_0^2-\omega^2}.$ (3 marks) Numbers: $\omega_0^2-\omega^2=4-1=3$, $\gamma\omega=1$. $|Z|=1/\sqrt{9+1}=1/\sqrt{10}\approx 0.3162$. $\arg(Z)=-\arctan(1/3)\approx -0.3217$ rad. (3 marks)

Question 2

(a) (5 marks) Roots are $1,\omega,\dots,\omega^{n-1}$ with $\omega=e^{2\pi i/n}\ne1$. Geometric series sum $=\frac{\omega^n-1}{\omega-1}=\frac{1-1}{\omega-1}=0$ (3). Geometric interpretation: the roots are equally spaced unit vectors (vertices of a regular $n$-gon centred at origin); by symmetry their vector sum is the centroid $\times n = 0$ (2).

(b) (4 marks) Product $=\prod_{k=0}^{n-1}e^{2\pi i k/n}=e^{2\pi i(0+1+\dots+(n-1))/n}=e^{2\pi i (n-1)/2}=e^{i\pi(n-1)}=(-1)^{n-1}=(-1)^{n+1}$. (4) (Alternatively: roots are roots of $z^n-1$; product $=(-1)^n(-1)=(-1)^{n+1}$ via constant term.)

(c) (7 marks) $-8+8\sqrt3 i$: modulus $=\sqrt{64+192}=\sqrt{256}=16$ (2). Argument: real $<0$, imag $>0$, second quadrant; $\tan^{-1}(8\sqrt3/8)=\tan^{-1}\sqrt3=\pi/3$, so argument $=\pi-\pi/3=2\pi/3$ (1). Thus $c=16e^{i(2\pi/3+2\pi k)}$. Fourth roots: modulus $16^{1/4}=2$ (1). Arguments $\theta_k=\frac{2\pi/3+2\pi k}{4}=\frac{\pi}{6}+\frac{\pi k}{2}$, $k=0,1,2,3$ (2): $\theta_0=\tfrac{\pi}{6},\ \theta_1=\tfrac{2\pi}{3},\ \theta_2=\tfrac{7\pi}{6}\to-\tfrac{5\pi}{6},\ \theta_3=\tfrac{5\pi}{3}\to-\tfrac{\pi}{3}.$ Roots: $2e^{i\pi/6},\,2e^{i2\pi/3},\,2e^{-i5\pi/6},\,2e^{-i\pi/3}$ (1, reduced to $(-\pi,\pi]$).

(d) (6 marks)

import numpy as np
def nth_roots(c, n):
    r = abs(c)**(1/n)             # modulus of each root
    phi = np.angle(c)             # principal argument of c
    return [r*np.exp(1j*(phi+2*np.pi*k)/n) for k in range(n)]

Explanation (3 marks): take $r=|c|^{1/n}$; base argument $\phi/n$; increment by $2\pi/n$ each time (equal spacing) to obtain all $n$ roots. Correct loop over $k=0..n-1$ (3 marks).

Question 3

(a) (6 marks) Divide $P$ by $(z-1)$: $z^3-5z^2+17z-13=(z-1)(z^2-4z+13)$ (2). Solve $z^2-4z+13=0$: $z=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm\sqrt{-36}}{2}=2\pm3i$ (3). These are conjugates $2+3i,\ 2-3i$ (1).

(b) (4 marks) If $P$ has real coefficients then $\overline{P(z)}=P(\bar z)$ (conjugation commutes with real-coefficient sums/products, 2). If $P(z_0)=0$ then $P(\bar z_0)=\overline{P(z_0)}=\bar0=0$, so $\bar z_0$ is also a root (2). Hence non-real roots pair up.

(c) (4 marks) Roots $1,\,2+3i,\,2-3i$. Sum $=1+4=5=-(-5)/1$ ✓ (1). Pairwise sum $=1(2+3i)+1(2-3i)+(2+3i)(2-3i)=4+13=17$ ✓ (2). Product $=1\cdot(4+9)=13=-(-13)/1$ ✓ (1).

(d) (4 marks) Required roots: $2+i,2-i,3-2i,3+2i$. $(z^2-4z+5)(z^2-6z+13)$ (2). Expand: $z^4-10z^3+42z^2-82z+65.$ (2)

[
 {"claim":"cos5θ expansion coefficients", "code":"import sympy as sp; th=sp.symbols('theta'); c=sp.cos(th); expr=sp.expand_trig(sp.cos(5*th)); target=16*c**5-20*c**3+5*c; result=sp.simplify(expr-target)==0"},
 {"claim":"|Z|=1/sqrt10 and arg=-atan(1/3)", "code":"D=(2**2-1**2)+sp.I*1*1; Z=1/D; result=(sp.simplify(sp.Abs(Z)-1/sp.sqrt(10))==0) and (sp.simplify(sp.arg(Z)+sp.atan(sp.Rational(1,3)))==0)"},
 {"claim":"fourth roots of -8+8sqrt3 i have modulus 2 and args pi/6,2pi/3,-5pi/6,-pi/3", "code":"c=-8+8*sp.sqrt(3)*sp.I; roots=sp.solve(sp.Symbol('z')**4-c, sp.Symbol('z')); mods={sp.nsimplify(sp.Abs(r)) for r in roots}; result=mods=={sp.Integer(2)}"},
 {"claim":"P factors and roots 1,2+3i,2-3i; quartic expansion", "code":"z=sp.symbols('z'); P=z**3-5*z**2+17*z-13; r=sp.solve(P,z); q=sp.expand((z**2-4*z+5)*(z**2-6*z+13)); result=(set(r)=={1,2+3*sp.I,2-3*sp.I}) and (q==z**4-10*z**3+42*z**2-82*z+65)"}
]