2010 H2 Mathematics Paper 1 Question 8

Complex Numbers

Answers

z1=2(cos13π+isin13π){z_1 = 2 \left( \cos \frac{1}{3} \pi + \mathrm{i} \sin \frac{1}{3} \pi \right)}
z2=2(cos(34π)+isin(34π)){z_2 = \sqrt{2} \left( \cos \left( - \frac{3}{4} \pi \right) + \mathrm{i} \sin \left( - \frac{3}{4} \pi \right) \right)}
(z1z2)=2(cos1112π+isin1112π){{\displaystyle \left(\frac{z_1}{z_2}\right)^*} = \sqrt{2} \left( \cos \frac{11}{12} \pi + \mathrm{i} \sin \frac{11}{12} \pi \right)}
Out of syllabus
Out of syllabus

Full solutions

(i)

z1=12+(3)2=2z2=(1)2+(1)2=2\begin{align*} \left| z_1 \right| &= \sqrt{1^2 + \left(\sqrt{3}\right)^2} \\ &= 2 \\ \left| z_2 \right| &= \sqrt{(-1)^2 + (-1)^2} \\ &= \sqrt{2} \\ \end{align*}
argz1=tan132=π3\begin{align*} \arg z_1 &= \tan^{-1} \frac{\sqrt{3}}{2} \\ &= \frac{\pi}{3} \end{align*}
For z2,{z_2,}
basic angle=tan111=π4argz2=(ππ4)=3π4\begin{align*} \textrm{basic angle} &= \tan^{-1} \frac{1}{1} \\ &= \frac{\pi}{4} \\ \arg z_2 &= -\left(\pi - \frac{\pi}{4} \right) \\ &= -\frac{3\pi}{4} \end{align*}
z1=2(cos13π+isin13π)  z2=2(cos(34π)+isin(34π))  \begin{align*} z_1 &= 2 \left( \cos \frac{1}{3} \pi + \mathrm{i} \sin \frac{1}{3} \pi \right) \; \blacksquare \\ z_2 &= \sqrt{2} \left( \cos \left( - \frac{3}{4} \pi \right) + \mathrm{i} \sin \left( - \frac{3}{4} \pi \right) \right) \; \blacksquare \end{align*}

(ii)

z1z2=2e13πi2e34πi=22ei(13π(34π))=2ei13π12=2e1112πi(z1z2)=2e1112πi=2(cos1112π+isin1112π)  \begin{align*} \frac{z_1}{z_2} &= \frac{2 \mathrm{e}^{\frac{1}{3} \pi \mathrm{i}}}{\sqrt{2} \mathrm{e}^{- \frac{3}{4} \pi \mathrm{i}}} \\ &= \frac{2}{\sqrt{2}} \mathrm{e}^{\mathrm{i}(\frac{1}{3} \pi-(- \frac{3}{4} \pi))}\\ &= \sqrt{2} \mathrm{e}^{\mathrm{i}\frac{13\pi}{12}}\\ &= \sqrt{2} \mathrm{e}^{- \frac{11}{12} \pi \mathrm{i}} \\ \left( \frac{z_1}{z_2} \right)^* &= \sqrt{2} \mathrm{e}^{\frac{11}{12} \pi \mathrm{i}} \\ &= \sqrt{2} \left( \cos \frac{11}{12} \pi + \mathrm{i} \sin \frac{11}{12} \pi \right) \; \blacksquare \end{align*}

(iii)

Out of syllabus

(iv)

Out of syllabus