2012 H2 Mathematics Paper 1 Question 6

Complex Numbers

Answers

(i)

(1−3c2)+i(3c−c3){(1 -3 c^2) + \mathrm{i}(3c-c^3)}

(ii)

1+i3{1+\mathrm{i}\sqrt{3}} or 1−i3{1-\mathrm{i}\sqrt{3}}

(iii)

Smallest n=10{n=10}
∣zn∣=1024{\left|z^n\right| = 1024}
arg⁡zn=2π3{\arg z^n = \frac{2\pi}{3}}

Full solutions

(i)

z3=(1+ic)3=(1+ic)2(1+ic)=(1+2ic−c2)(1+ic)=1+2ic−c2+ic−2c2−ic3=(1−3c2)+i(3c−c3)  ■\begin{align*} z^3 &= (1+\mathrm{i}c)^3 \\ &= (1+\mathrm{i}c)^2(1+\mathrm{i}c) \\ &= (1+2\mathrm{i}c - c^2)(1+\mathrm{i}c) \\ &= 1+2\mathrm{i}c - c^2 + \mathrm{i}c - 2c^2 - \mathrm{i}c^3 \\ &= (1 -3 c^2) + \mathrm{i}(3c-c^3) \; \blacksquare \end{align*}

(ii)

Im(z3)=03c−c3=0c(3−c2)=0c=±3   or   c=0(NA)\begin{gather*} \textrm{Im}(z^3) = 0 \\ 3c-c^3 = 0 \\ c(3-c^2) = 0 \\ c = \pm \sqrt{3} \; \textrm{ or } \; c = 0 \textrm{(NA)} \end{gather*}
z=1+i3   or   z=1−i3  ■z=1+\mathrm{i}\sqrt{3} \; \textrm{ or } \; z=1-\mathrm{i}\sqrt{3} \; \blacksquare

(iii)

z=1−i3z=1-\mathrm{i}\sqrt{3}
∣zn∣>1000∣z∣n>100012+(3)2>10002n>1000nln⁡2>ln⁡1000n>9.9658\begin{align*} \left| z^n \right | &> 1000 \\ \left| z \right |^n &> 1000 \\ \sqrt{1^2 + \left(\sqrt{3}\right)^2} &> 1000 \\ 2^n &> 1000 \\ n \ln 2 &> \ln 1000 \\ n &> 9.9658 \end{align*}
Hence the smallest positive integer n=10  ■{n = 10 \; \blacksquare}
∣zn∣=210=1024  ■\begin{align*} \left| z^n \right | &= 2^{10} \\ &= 1024 \; \blacksquare \end{align*}
basic angle=tan⁡−13=π3arg⁡z=−π3\begin{align*} \textrm{basic angle} &= \tan^{-1} \sqrt{3} \\ &= \frac{\pi}{3} \\ \arg z &= -\frac{\pi}{3} \end{align*}
arg⁡zn=10arg⁡z+2kπ=10(−π3)+4π=2π3  ■\begin{align*} \arg z^{n} &= 10 \arg z + 2k\pi \\ &= 10\left(-\frac{\pi}{3}\right) + 4\pi \\ &= \frac{2\pi}{3} \; \blacksquare \end{align*}