Question

Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
(i) 1 + i
(ii) $\sqrt{3}+i$
(iii) 1 − i
(iv) $\frac{1-i}{1+i}$
(v) $\frac{1}{1+i}$
(vi) $\frac{1+2i}{1-3i}$
(vii) $\sin 120°-i\cos 120°$
(viii) $\frac{-16}{1+i\sqrt{3}}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)z=1+i \\ r=\left|z\right| \\ =\sqrt{1+1} \\ =\sqrt{2} \\ \mathrm{Let}\tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \Rightarrow \tan \alpha =\left(\frac{1}{1}\right) \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{4} \\ \mathrm{Since}\mathrm{point}(1,1)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant},\mathrm{the}\mathrm{argument}\mathrm{of}z\mathrm{is}\mathrm{given}\mathrm{by} \\ \theta =\alpha =\frac{\mathrm{\pi }}{4} \\ \mathrm{Polar}\mathrm{form}=r\left(\cos \theta +i\sin \theta \right) \\ =\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{4}+i\sin \frac{\mathrm{\pi }}{4}\right) \\ \left(\mathrm{ii}\right)z=\sqrt{3}+i \\ r=\left|z\right| \\ =\sqrt{3+1} \\ =\sqrt{4} \\ =2 \\ \mathrm{Let}\tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \Rightarrow \tan \alpha =\left(\frac{1}{\sqrt{3}}\right) \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{6} \\ \mathrm{Since}\mathrm{point}(\sqrt{3},1)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant},\mathrm{the}\mathrm{argument}\mathrm{of}z\mathrm{is}\mathrm{given}\mathrm{by} \\ \theta =\alpha =\frac{\mathrm{\pi }}{6} \end{gathered}$$

$$\begin{gathered} \mathrm{Polar}\mathrm{form}=r\left(\cos \theta +i\sin \theta \right) \\ =2\left(\cos \frac{\mathrm{\pi }}{6}+i\sin \frac{\mathrm{\pi }}{6}\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)z=1-i \\ r=\left|z\right| \\ =\sqrt{1+1} \\ =\sqrt{2} \\ \mathrm{Let}\tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \therefore \tan \alpha =\left|\frac{-1}{1}\right| \\ =\frac{\mathrm{\pi }}{4} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{4} \\ \mathrm{Since}\mathrm{point}(1,-1)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{fourth}\mathrm{quadrant},\mathrm{the}\mathrm{argument}\mathrm{of}z\mathrm{is}\mathrm{given}\mathrm{by} \\ \theta =-\alpha =-\frac{\mathrm{\pi }}{4} \\ \mathrm{Polar}\mathrm{form}=r\left(\cos \theta +i\sin \theta \right) \\ =\sqrt{2}\left\{\cos \left(-\frac{\mathrm{\pi }}{4}\right)+i\sin \left(-\frac{\mathrm{\pi }}{4}\right)\right\} \\ =\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{4}-i\sin \frac{\mathrm{\pi }}{4}\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\frac{1-i}{1+i} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator}: \\ \frac{1-i}{1+i}\times \frac{1-i}{1-i} \\ \Rightarrow \frac{1+i^2-2i}{1-i^2} \\ \Rightarrow \frac{-2i}{2}\left(\because i^2=-1\right) \\ \Rightarrow -i \\ r=\left|z\right| \\ =\sqrt{0+1} \\ =1 \\ \mathrm{Since}\mathrm{point}(0,-1)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{negative}\mathrm{direction}\mathrm{of}\mathrm{the}\mathrm{imaginary}\mathrm{axis},\mathrm{the}\mathrm{argument}\mathrm{of}z\mathrm{is}\mathrm{given}\mathrm{by}\frac{3\mathrm{\pi }}{2}. \\ \mathrm{Polar}\mathrm{form}=r\left(\cos \theta +i\sin \theta \right) \\ =\left(c\mathrm{os}\frac{3\mathrm{\pi }}{2}+i\sin \frac{3\mathrm{\pi }}{2}\right) \\ =\left\{c\mathrm{os}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{2}\right)+i\sin \left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{2}\right)\right\} \\ =\left(\cos \frac{\mathrm{\pi }}{2}-i\sin \frac{\mathrm{\pi }}{2}\right) \end{gathered}$$

$$\begin{gathered} \left(v\right)\frac{1}{1+i} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator}: \\ \frac{1}{1+i}\times \frac{1-i}{1-i} \\ \Rightarrow \frac{1-i}{1-i^2} \\ \Rightarrow \frac{1-i}{2}\left(\because i^2=-1\right) \\ \Rightarrow \frac{1}{2}-\frac{i}{2} \\ r=\left|z\right| \\ =\sqrt{\frac{1}{4}+\frac{1}{4}} \\ =\frac{1}{\sqrt{2}} \\ \mathrm{Let}\tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \therefore \tan \alpha =\left|\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{-1}{2}}}\right| \\ =1 \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{4} \\ \mathrm{Since}\mathrm{point}\left(\frac{1}{2},-\frac{1}{2}\right)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{fourth}\mathrm{quadrant},\mathrm{the}\mathrm{argument}\mathrm{is}\mathrm{given}\mathrm{by} \\ \theta =-\alpha =\frac{-\mathrm{\pi }}{4} \\ \mathrm{Polar}\mathrm{form}=r\left(\cos \theta +i\sin \theta \right) \\ =\frac{1}{\sqrt{2}}\left\{c\mathrm{os}\left(\frac{-\mathrm{\pi }}{4}\right)+i\sin \left(\frac{-\mathrm{\pi }}{4}\right)\right\} \\ =\frac{1}{\sqrt{2}}\left(c\mathrm{os}\frac{\mathrm{\pi }}{4}-i\sin \frac{\mathrm{\pi }}{4}\right) \end{gathered}$$

$$\begin{gathered} \left(vi\right)\frac{1+2i}{1-3i} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator}: \\ \frac{1+2i}{1-3i}\times \frac{1+3i}{1+3i} \\ \Rightarrow \frac{1+3i+2i+6i^2}{1-9i^2} \\ \Rightarrow \frac{-5+5i}{10}\left(\because i^2=-1\right) \\ \Rightarrow \frac{-1}{2}+\frac{i}{2} \\ r=\left|z\right| \\ =\sqrt{\frac{1}{4}+\frac{1}{4}} \\ =\frac{1}{\sqrt{2}} \\ \mathrm{Let}\tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \mathrm{Then},\tan \alpha =\left|\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{-1}{2}}}\right| \\ =1 \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{4} \\ \mathrm{Since}\mathrm{point}\left(\frac{-1}{2},\frac{1}{2}\right)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{quadrant},\mathrm{the}\mathrm{argument}\mathrm{is}\mathrm{given}\mathrm{by} \\ \theta =\mathrm{\pi }-\alpha \\ =\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ =\frac{3\mathrm{\pi }}{4} \\ \mathrm{Polar}\mathrm{form}=r\left(\cos \theta +i\sin \theta \right) \\ =\frac{1}{\sqrt{2}}\left(c\mathrm{os}\frac{3\mathrm{\pi }}{4}+i\sin \frac{3\mathrm{\pi }}{4}\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right)\sin 120°-i\cos 120° \\ \frac{\sqrt{3}}{2}+\frac{i}{2} \\ r=\left|z\right| \\ =\sqrt{\frac{3}{4}+\frac{1}{4}} \\ =1 \\ \mathrm{Let}\tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \mathrm{Then},\tan \alpha =\left|\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right| \\ =\frac{1}{\sqrt{3}} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{6} \\ \mathrm{Since}\mathrm{point}\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant},\mathrm{the}\mathrm{argument}\mathrm{is}\mathrm{given}\mathrm{by} \\ \theta =\alpha =\frac{\mathrm{\pi }}{6} \\ \mathrm{Polar}\mathrm{form}=r\left(\cos \theta +i\sin \theta \right) \\ =\cos \frac{\mathrm{\pi }}{6}+i\sin \frac{\mathrm{\pi }}{6} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{viii}\right)\frac{-16}{1+i\sqrt{3}} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator}: \\ \frac{-16}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}} \\ \Rightarrow \frac{-16+16\sqrt{3}i}{1-3i^2} \\ \Rightarrow \frac{-16+16\sqrt{3}i}{4}\left(\because i^2=-1\right) \\ \Rightarrow -4+4\sqrt{3}i \\ r=\left|z\right| \\ =\sqrt{16+48} \\ =8 \\ \mathrm{Let}\tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \mathrm{Then},\tan \alpha =\left|\frac{{\displaystyle 4\sqrt{3}}}{-{\displaystyle 4}}\right| \\ =\sqrt{3} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{3} \\ \mathrm{Since}\mathrm{the}\mathrm{point}\left(-4,4\sqrt{3}\right)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{quadrant},\mathrm{the}\mathrm{argument}\mathrm{is}\mathrm{given}\mathrm{by} \\ \theta =\mathrm{\pi }-\alpha \\ =\mathrm{\pi }-\frac{\mathrm{\pi }}{3} \\ =\frac{2\mathrm{\pi }}{3} \\ \mathrm{Polar}\mathrm{form}=r\left(\cos \theta +i\sin \theta \right) \\ =8\left\{c\mathrm{os}\frac{2\mathrm{\pi }}{3}+i\sin \frac{2\mathrm{\pi }}{3}\right\} \end{gathered}$$