2023 H2 Mathematics Paper 1 Question 8

Complex Numbers

Answers

(a)

z=2e23πi.{{ z = 2 \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}}.}}

(b)

Smallest positive integer n=2.{{ n = 2.}}

(c)

w=−4+3i,v=−2{{w = - 4 + 3 \mathrm{i}, v = - 2}}
or w=−4+215i,v=−125.{{\quad w = - 4 + \frac{21}{5} \mathrm{i}, v = - \frac{12}{5}.}}

Full solutions

(a)

∣z∣=12+(3)2=2arg⁡(z)=π−tan⁡−13=23πz=2e23πi  ■\begin{align*} |z| &= \sqrt{ 1^2 + (\sqrt{3})^2 } \\ &= 2 \\ \arg (z) &= \pi - \tan^{-1} \sqrt{3} \\ &= \frac{2}{3} \pi \\ z &= 2 \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}} \; \blacksquare \end{align*}

(b)

zniz∗=(2e23πi)ne12πi⋅2e−23πi=2n−1e⁡(2n3π+16π)i\begin{align*} & \frac{z^n}{\mathrm{i} z^*} \\ & = \frac{\left(2 \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}}\right)^n}{\mathrm{e}^{\frac{1}{2} \pi \mathrm{i}} \cdot 2 \mathrm{e}^{- \frac{2}{3} \pi \mathrm{i}}} \\ & = 2^{n-1} \operatorname{e}^{\left( \frac{2n}{3}\pi + \frac{1}{6} \pi \right)\mathrm{i}} \end{align*}

For the complex number to be purely imaginary, for k∈Z,{{k \in \mathbb{Z},}}

2n3π−16π=π2+kπ2n3π=13π+kπn=1+3k2\begin{align*} \frac{2n}{3}\pi - \frac{1}{6} \pi &= \frac{\pi}{2} + k \pi \\ \frac{2n}{3}\pi &= \frac{1}{3} \pi + k \pi \\ n &= \frac{1+3k}{2} \end{align*}

Smallest positive integer value of n{{n}} occurs when k=1{{k = 1}}

n=2  ■{n = 2 \; \blacksquare}

(c)

2v+∣w∣=13v−iw=−3+4i\begin{align} 2v + |w| &= 1 \\ 3v - \mathrm{i}w &= - 3 + 4 \mathrm{i} \end{align}

From (2),{{(2),}}

v=−3+4i+iw3\begin{equation} v = \frac{- 3 + 4 \mathrm{i} + \mathrm{i}w}{3} \end{equation}

Substituting into (1),{{(1),}}

2(−3+4i+iw3)+∣w∣=1{ 2 \left( \frac{- 3 + 4 \mathrm{i} + \mathrm{i}w}{3} \right) + |w| = 1}

Let w=x+yi{{w = x+y\mathrm{i}}}

2(−3+4i+i(x+yi)3)+x2+y2=1−6+8i+2xi−2y+3x2+y2=3\begin{align*} 2 \left( \frac{- 3 + 4 \mathrm{i} + \mathrm{i}(x+yi)}{3} \right) + \sqrt{x^2 + y^2} &= 1 \\ - 6 + 8 \mathrm{i} + 2x \mathrm{i} - 2y + 3 \sqrt{x^2 + y^2} &= 3 \end{align*}

Comparing imaginary parts,

8+2x=02x=−8x=−4\begin{align*} 8 + 2 x &= 0 \\ 2 x &= - 8 \\ x &= - 4 \end{align*}

Substituting x=−4{{x=- 4}} and comparing real parts,

−6−2y+3(−4)2+y2=3316+y2=9+2y9(16+y2)=(9+2y)29y2+144=4y2+36y+815y2−36y+63=0(y−3)(5y−21)=0\begin{align*} -6 - 2y + 3 \sqrt{(- 4)^2 + y^2} &= 3 \\ 3 \sqrt{16 + y^2} &= 9 + 2y \\ 9 (16 + y^2) &= (9 + 2y)^2 \\ 9 y^2 + 144 &= 4 y^2 + 36 y + 81 \\ 5 y^2 - 36 y + 63 &= 0 \\ \left( y - 3 \right) \left( 5 y - 21 \right) &= 0 \end{align*}
y=3ory=215w=−4+3iorw=−4+215i\begin{gather*} y = 3 \quad \text{or} \quad y = \frac{21}{5} \\ w = - 4 + 3 \mathrm{i} \quad \text{or} \quad w = - 4 + \frac{21}{5} \mathrm{i} \end{gather*}

Substituting w{{w}} into (1),{{(1),}}

v=−3+4i+i(−4+3i)3orv=−3+4i+i(−4+215i)3v=−2v=−125\begin{alignat*}{3} v &= \frac{- 3 + 4 \mathrm{i}+\mathrm{i}(- 4 + 3 \mathrm{i})}{3} & \quad &\quad \text{or} \quad \quad & v &= \frac{- 3 + 4 \mathrm{i}+\mathrm{i}(- 4 + \frac{21}{5} \mathrm{i})}{3} \\ v &= - 2 &&& v &= - \frac{12}{5} \end{alignat*}
w=−4+3i,v=−2  ■orw=−4+215i,v=−125  ■\begin{align*} & w = - 4 + 3 \mathrm{i}, \quad v = - 2 \; \blacksquare \\ \quad \text{or} \quad & w = - 4 + \frac{21}{5} \mathrm{i}, \quad v = - \frac{12}{5} \; \blacksquare \end{align*}