Question

Express the following complex in the form r(cos θ + i sin θ):
(i) 1 + i tan α
(ii) tan α − i
(iii) 1 − sin α + i cos α

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}\mathrm{z}=1+i\tan \alpha \mathit{} \\ \mathit{\because }\tan \alpha \mathit{}\mathrm{is}\mathrm{periodic}\mathrm{with}\mathrm{period}\mathit{}\mathrm{\pi }.\mathrm{So},\mathrm{let}\mathrm{us}\mathrm{take} \\ \alpha \in [0,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.)\cup (\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\mathrm{\pi }] \\ \mathrm{Case}\mathrm{I}: \\ \mathrm{When}\alpha \in [0,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.) \\ z=1+i\tan \mathrm{\alpha } \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{1+{\tan }^2\alpha } \\ =\left|\sec \alpha \right|\left[\because 0<\alpha <\frac{\mathrm{\pi }}{2}\right] \\ =\sec \alpha \\ \mathrm{Let}\beta \mathrm{be}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{given}\mathrm{by}\tan \beta =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \tan \beta =\left|\tan \mathit{}\alpha \right| \\ =\tan \alpha \\ \Rightarrow \beta =\alpha \\ \mathrm{As}z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}.\mathrm{Therefore},\arg \left(z\right)=\beta =\alpha \\ \mathrm{Thus},z\mathrm{in}\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{is}\mathrm{given}\mathrm{by} \\ z=\sec \alpha \left(\cos \alpha +i\sin \alpha \right) \\ \mathrm{Case}\mathrm{II}: \\ z=1+i\tan \mathit{}\alpha \\ \Rightarrow \left|z\right|=\sqrt{1+{\tan }^2\alpha } \\ =\left|\sec \mathrm{\alpha }\right|\left[\because \frac{\mathrm{\pi }}{2}<\alpha <\mathrm{\pi }\right] \\ =-\sec \mathrm{\alpha } \\ \mathrm{Let}\beta \mathit{}\mathrm{be}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{given}\mathrm{by}\tan \beta =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\mathit{\right)}}\right| \\ \tan \beta =\left|\tan \alpha \right| \\ =-\tan \mathit{}\alpha \\ \Rightarrow \tan \beta =\tan \left(\mathrm{\pi }-\mathrm{\alpha }\right) \\ \Rightarrow \beta =\mathrm{\pi }-\mathrm{\alpha } \\ \mathrm{As},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{fourth}\mathrm{quadrant}. \\ \therefore \arg \left(z\right)=-\beta =\alpha -\mathrm{\pi } \\ \mathrm{Thus},z\mathrm{in}\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{is}\mathrm{given}\mathrm{by} \\ z=-\sec \alpha \left\{\cos \left(\alpha -\mathrm{\pi }\right)+i\sin \left(\alpha -\mathrm{\pi }\right)\right\} \end{gathered}$$



$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let}\mathrm{z}=\tan \alpha -i\mathit{} \\ \mathit{\because }\tan \alpha \mathit{}\mathrm{is}\mathrm{periodic}\mathrm{with}\mathrm{period}\mathit{}\mathrm{\pi }.\mathrm{So},\mathrm{let}\mathrm{us}\mathrm{take} \\ \alpha \in [0,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.)\cup (\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\mathrm{\pi }] \\ \mathrm{Case}\mathrm{I}: \\ z=\tan \mathit{}\alpha -\mathrm{i} \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{{\tan }^2+1} \\ =\left|\sec \mathrm{\alpha }\right|\left[\because 0<\mathrm{\alpha }<\frac{\mathrm{\pi }}{2}\right] \\ =\sec \mathrm{\alpha } \\ \mathrm{Let}\mathrm{\beta }\mathrm{be}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{given}\mathrm{by}\tan \mathrm{\beta }=\left|\frac{\mathrm{Im}\left(\mathrm{z}\right)}{\mathrm{Re}\left(\mathrm{z}\right)}\right| \\ \tan \beta =\frac{1}{\left|\tan \mathit{}\alpha \right|} \\ =\left|\cot \mathit{}\alpha \right| \\ =\cot \alpha \\ =\tan \left(\frac{\mathrm{\pi }}{2}-\alpha \right) \\ \Rightarrow \beta =\frac{\mathrm{\pi }}{2}-\alpha \\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{Re}\left(z\right)>0\mathrm{and}\mathrm{Im}\left(z\right)<0.\mathrm{So},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{fourth}\mathrm{quadrant}. \\ \therefore \arg \left(z\right)=-\beta =\alpha -\frac{\mathrm{\pi }}{2} \\ \mathrm{Thus},z\mathrm{in}\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{is}\mathrm{given}\mathrm{by} \\ z=\sec \alpha \left\{\cos \left(\alpha -\frac{\mathrm{\pi }}{2}\right)+i\sin \left(\alpha -\frac{\mathrm{\pi }}{2}\right)\right\} \\ \mathrm{Case}\mathrm{II}: \\ z=\tan \mathit{}\alpha -i \\ \Rightarrow \left|z\right|=\sqrt{{\tan }^2+1} \\ =\left|\sec \mathrm{\alpha }\right|\left[\because \frac{\mathrm{\pi }}{2}<\alpha <\mathrm{\pi }\right] \\ =-\sec \mathrm{\alpha } \\ \mathrm{Let}\beta \mathit{}\mathrm{be}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{given}\mathrm{by}\tan \beta =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\mathit{\right)}}\right| \\ \tan \beta =\frac{1}{\left|\tan \mathit{}\alpha \right|} \\ =\left|\cot \alpha \right| \\ =-\cot \mathit{}\alpha \\ =\tan \left(\alpha -\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \beta =\alpha -\frac{\mathrm{\pi }}{2} \\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{R}e\left(z\right)<0\mathrm{and}\mathrm{Im}\left(z\right)<0.\mathrm{So},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{quadrant}. \\ \therefore \arg \left(z\right)=\mathrm{\pi }+\beta =\frac{\mathrm{\pi }}{2}+\alpha \\ \mathrm{Thus},z\mathrm{in}\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{is}\mathrm{given}\mathrm{by} \\ z=-\sec \alpha \left\{\cos \left(\frac{\mathrm{\pi }}{2}+\alpha \right)+i\sin \left(\frac{\mathrm{\pi }}{2}+\alpha \right)\right\} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let}\mathrm{z}=\left(1-\sin \alpha \right)+i\cos \alpha . \\ \mathit{\because }\mathrm{sine}\mathrm{and}\mathrm{cosine}\mathrm{functions}\mathrm{are}\mathrm{periodic}\mathrm{functions}\mathrm{with}\mathrm{period}2\mathrm{\pi }. \\ \mathrm{So},\mathrm{let}\mathrm{us}\mathrm{take}\alpha \in [0,2\mathrm{\pi }] \\ \mathrm{Now},z=1-\sin \alpha +i\cos \alpha \\ \Rightarrow \left|z\right|=\sqrt{{\left(1-\sin \alpha \right)}^2+{\cos }^2\alpha }=\sqrt{2-\sin \alpha }=\sqrt{2}\sqrt{1-\sin \alpha } \\ \Rightarrow \left|z\right|=\sqrt{2}\sqrt{{\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right)}^2}=\sqrt{2}\left|\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right| \\ \mathrm{Let}\beta \mathrm{be}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{given}\mathrm{by}\tan \beta =\frac{\left|\mathrm{Im}\left(z\right)\right|}{\left|\mathrm{Re}\left(z\right)\right|}.\mathrm{Then}, \\ \tan \beta =\frac{\left|\cos \alpha \right|}{\left|1-\sin \alpha \right|}=\left|\frac{{\cos }^2\frac{\alpha }{2}-{\sin }^2\frac{\alpha }{2}}{{\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right)}^2}\right|=\left|\frac{\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}}\right| \\ \Rightarrow \tan \beta =\left|\frac{1+\tan \frac{\alpha }{2}}{1-\tan \frac{\alpha }{2}}\right|=\left|\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)\right| \\ \mathrm{Case}\mathrm{I}:\mathrm{When}0\le \alpha <\frac{\mathrm{\pi }}{2} \\ \mathrm{In}\mathrm{this}\mathrm{case},\mathrm{we}\mathrm{have}, \\ \cos \frac{\alpha }{2}>\sin \frac{\alpha }{2}\mathrm{and}\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\in [\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}) \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{2}\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right) \\ \mathrm{and}\tan \beta =\left|\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)\right|=\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right) \\ \Rightarrow \beta =\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2} \\ \mathrm{Clearly},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}.\mathrm{Therefore},\arg \left(z\right)=\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{of}z\mathrm{is} \\ \sqrt{2}\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right)\left\{\cos \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)+i\sin \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)\right\} \\ \mathrm{Case}\mathrm{II}:\mathrm{When}\frac{\mathrm{\pi }}{2}<\alpha <\frac{3\mathrm{\pi }}{2} \\ \mathrm{In}\mathrm{this}\mathrm{case},\mathrm{we}\mathrm{have}, \\ \cos \frac{\alpha }{2}<\sin \frac{\alpha }{2}\mathrm{and}\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right) \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{2}\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right)=-\sqrt{2}\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right) \\ \mathrm{and}\tan \beta =\left|\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)\right|=-\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)=\tan \left\{\mathrm{\pi }-\left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)\right\}=\tan \left(\frac{3\mathrm{\pi }}{4}-\frac{\alpha }{2}\right) \\ \Rightarrow \beta =\frac{3\mathrm{\pi }}{4}-\frac{\alpha }{2} \\ \mathrm{Clearly},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{fourth}\mathrm{quadrant}.\mathrm{Therefore},\arg \left(z\right)=-\beta =-\left(\frac{3\mathrm{\pi }}{4}-\frac{\alpha }{2}\right)=\frac{\alpha }{2}-\frac{3\mathrm{\pi }}{4} \\ \mathrm{Hence},\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{of}z\mathrm{is} \\ -\sqrt{2}\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right)\left\{\cos \left(\frac{\alpha }{2}-\frac{3\mathrm{\pi }}{4}\right)+i\sin \left(\frac{\alpha }{2}-\frac{3\mathrm{\pi }}{4}\right)\right\} \\ \mathrm{Case}\mathrm{III}:\mathrm{When}\frac{3\mathrm{\pi }}{2}<\alpha <2\mathrm{\pi } \\ \mathrm{In}\mathrm{this}\mathrm{case},\mathrm{we}\mathrm{have}, \\ \cos \frac{\mathrm{\alpha }}{2}<\sin \frac{\mathrm{\alpha }}{2}\mathrm{and}\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\alpha }}{2}\in \left(\mathrm{\pi },\frac{5\mathrm{\pi }}{4}\right) \\ \Rightarrow \left|\mathrm{z}\right|=\sqrt{2}\left|\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right|=-\sqrt{2}\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right) \\ \mathrm{and}\tan \beta =\left|\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)\right|=\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)=-\tan \left\{\mathrm{\pi }-\left(\frac{\mathrm{\pi }}{4}+\frac{\alpha }{2}\right)\right\}=\tan \left(\frac{\alpha }{2}-\frac{3\mathrm{\pi }}{4}\right) \\ \Rightarrow \beta =\frac{\alpha }{2}-\frac{3\mathrm{\pi }}{4} \\ \mathrm{Clearly},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}.\mathrm{Therefore},\arg \left(z\right)=\beta =\frac{\alpha }{2}-\frac{3\mathrm{\pi }}{4} \\ \mathrm{Hence},\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{of}z\mathrm{is} \\ -\sqrt{2}\left(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\right)\left\{\cos \left(\frac{\alpha }{2}-\frac{3\mathrm{\pi }}{4}\right)+i\sin \left(\frac{\alpha }{2}-\frac{3\mathrm{\pi }}{4}\right)\right\} \end{gathered}$$