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Trigonometric Ratios of Multiples of an Angle

Express the f...

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
(iv) $\frac{1 - i}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$

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Solution

$(i) Let z = 1 + i \tan α ∵ \tan α is periodic with period π . So, let us take α \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π] Case I : When α \in [0, \frac{π}{2}) z = 1 + i \tan α \Rightarrow |z| = \sqrt{1 + \tan^{2} α} = |\sec α| [∵ 0 < α < \frac{π}{2}] = \sec α Let β be an acute angle given by \tan β = |\frac{Im (z)}{Re (z)}| \tan β = |\tan α| = \tan α \Rightarrow β = α As z lies in the first quadrant . Therefore, \arg (z) = β = α Thus, z in the polar form is given by z = \sec α (\cos α + i \sin α) Case II : z = 1 + i \tan α \Rightarrow |z| = \sqrt{1 + \tan^{2} α} = |\sec α| [∵ \frac{π}{2} < α < π] = - \sec α Let β be an acute angle given by \tan β = |\frac{Im (z)}{Re (z)}| \tan β = |\tan α| = - \tan α \Rightarrow \tan β = \tan (π - α) \Rightarrow β = π - α As, z lies in the fourth quadrant . ∴ \arg (z) = - β = α - π Thus, z in the polar form is given by z = - \sec α \{\cos (α - π) + i \sin (α - π)\}$

$(ii) Let z = \tan α - i ∵ \tan α is periodic with period π . So, let us take α \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π] Case I : z = \tan α - i \Rightarrow |z| = \sqrt{\tan^{2} + 1} = |\sec α| [∵ 0 < α < \frac{π}{2}] = \sec α Let β be an acute angle given by \tan β = |\frac{Im (z)}{Re (z)}| \tan β = \frac{1}{|\tan α|} = |\cot α| = \cot α = \tan (\frac{π}{2} - α) \Rightarrow β = \frac{π}{2} - α We can see that Re (z) > 0 and Im (z) < 0 . So, z lies in the fourth quadrant . ∴ \arg (z) = - β = α - \frac{π}{2} Thus, z in the polar form is given by z = \sec α \{\cos (α - \frac{π}{2}) + i \sin (α - \frac{π}{2})\} Case II : z = \tan α - i \Rightarrow |z| = \sqrt{\tan^{2} + 1} = |\sec α| [∵ \frac{π}{2} < α < π] = - \sec α Let β be an acute angle given by \tan β = |\frac{Im (z)}{Re (z)}| \tan β = \frac{1}{|\tan α|} = |\cot α| = - \cot α = \tan (α - \frac{π}{2}) \Rightarrow β = α - \frac{π}{2} We can see that R e (z) < 0 and Im (z) < 0 . So, z lies in the third quadrant . ∴ \arg (z) = π + β = \frac{π}{2} + α Thus, z in the polar form is given by z = - \sec α \{\cos (\frac{π}{2} + α) + i \sin (\frac{π}{2} + α)\}$

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

$(iv) Let z = \frac{1 - i}{c o s \frac{π}{3} + i s i n \frac{π}{3}} = \frac{1 - i}{\frac{1}{2} + i \frac{\sqrt{3}}{2}} = \frac{2 - 2 i}{1 + i \sqrt{3}} \times \frac{1 - i \sqrt{3}}{1 - i \sqrt{3}} = \frac{2 - 2 i - 2 \sqrt{3} i + 2 \sqrt{3} i^{2}}{1 + 3} = \frac{2 - 2 \sqrt{3} - 2 i (1 + \sqrt{3})}{4} = \frac{(1 - \sqrt{3}) + i (- 1 - \sqrt{3})}{2} = \frac{(1 - \sqrt{3})}{2} + i \frac{(- 1 - \sqrt{3})}{2} Now, z = \frac{(1 - \sqrt{3})}{2} + i \frac{(- 1 - \sqrt{3})}{2} \Rightarrow |z| = \sqrt{{(\frac{1 - \sqrt{3}}{2})}^{2} + {(\frac{- 1 - \sqrt{3}}{2})}^{2}} = \sqrt{(\frac{1 + 3 - 2 \sqrt{3}}{4}) + (\frac{1 + 3 + 2 \sqrt{3}}{4})} = \sqrt{\frac{8}{4}} = \sqrt{2} Let β be an acute angle given by \tan β = \frac{|Im (z)|}{|Re (z)|} . Then, \tan β = \frac{|\frac{1 + \sqrt{3}}{2}|}{|\frac{1 - \sqrt{3}}{2}|} = |\frac{1 + \sqrt{3}}{1 - \sqrt{3}}| = |\frac{\tan \frac{π}{4} + \tan \frac{π}{3}}{1 - \tan \frac{π}{4} \tan \frac{π}{3}}| \Rightarrow \tan β = |\tan (\frac{π}{4} + \frac{π}{3})| = |\tan \frac{7 π}{12}| \Rightarrow β = \frac{7 π}{12} Clearly, z lies in the fourth quadrant . Therefore, \arg (z) = - \frac{7 π}{12} Hence, the polar form of z is \sqrt{2} (\cos \frac{7 π}{12} - \sin \frac{7 π}{12})$

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