Question Number 43363 by rahul 19 last updated on 10/Sep/18 | ||
$$\mathrm{If}\:\mathrm{z}=\:\mathrm{cos}\:\theta\:+\:\mathrm{isin}\:\theta\:,\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{6}}\:,\:\mathrm{then}\:\mathrm{prove} \\ $$ $$\mathrm{that}\:\mathrm{argument}\:\mathrm{of}\:\:\mathrm{1}−\mathrm{z}^{\mathrm{4}} \:=\:\mathrm{2}\theta\:−\:\frac{\pi}{\mathrm{2}}\:. \\ $$ | ||
Commented bymaxmathsup by imad last updated on 10/Sep/18 | ||
$${we}\:{have}\:{z}\:={e}^{{i}\theta} \:\Rightarrow\mathrm{1}−{z}^{\mathrm{4}} \:=\mathrm{1}−{e}^{\mathrm{4}{i}\theta} \:\:\:=\mathrm{1}−{cos}\left(\mathrm{4}\theta\right)−{isin}\left(\mathrm{4}\theta\right) \\ $$ $$=\mathrm{2}\:{sin}^{\mathrm{2}} \left(\mathrm{2}\theta\right)\:−\mathrm{2}{isin}\left(\mathrm{2}\theta\right){cos}\left(\mathrm{2}\theta\right)\:=−\mathrm{2}{isin}\left(\mathrm{2}\theta\right)\left\{{cos}\left(\mathrm{2}\theta\right)\:+{isin}\left(\mathrm{2}\theta\right)\right\} \\ $$ $$=\mathrm{2}\:{sin}\left(\mathrm{2}\theta\right)\left(−\boldsymbol{{i}}\right)\:\boldsymbol{{e}}^{\boldsymbol{{i}}\left(\mathrm{2}\theta\right)} \:\:{we}\:{have}\:\:\mathrm{0}<\mathrm{2}\theta<\frac{\pi}{\mathrm{3}}\:\Rightarrow{sin}\left(\mathrm{2}\theta\right)>\mathrm{0}\:\Rightarrow\mid\mathrm{1}−{z}^{\mathrm{4}} \mid\:=\mathrm{2}\:{sin}\left(\mathrm{2}\theta\right) \\ $$ $${and}\:\:{arg}\left(\mathrm{1}−{z}^{\mathrm{4}} \right)\:\equiv{arg}\left(−{i}\right)\:+\mathrm{2}\theta\:\:\left[\mathrm{2}\pi\right]\:\Rightarrow{arg}\left(\mathrm{1}−{z}^{\mathrm{4}} \right)\equiv\:−\frac{\pi}{\mathrm{2}}\:+\mathrm{2}\theta\left[\mathrm{2}\pi\right]\:. \\ $$ | ||
Answered by ajfour last updated on 10/Sep/18 | ||
$$\mathrm{1}−{z}^{\mathrm{4}} =\mathrm{1}−\left({x}+{iy}\right)^{\mathrm{4}} \\ $$ $$\:\:=\:\left(\mathrm{1}−{x}^{\mathrm{4}} −{y}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)+\mathrm{4}{i}\left({xy}^{\mathrm{3}} −{x}^{\mathrm{3}} {y}\right) \\ $$ $$\:\:\mathrm{tan}\:\phi\:=\:\frac{\mathrm{4}{xy}\left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)}{\mathrm{1}−\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$ $$\:\:\:\:=\:\frac{\mathrm{4}{m}\left({m}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{1}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} } \\ $$ $$\:\:\:\:\left({if}\:\:{m}=\mathrm{tan}\:\theta\:\right) \\ $$ $$\Rightarrow\:\:\:\mathrm{tan}\:\phi\:=\:\frac{{m}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{m}}\:=\:−\mathrm{cot}\:\mathrm{2}\theta \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{tan}\:\left(\mathrm{2}\theta−\frac{\pi}{\mathrm{2}}\right) \\ $$ $$\Rightarrow\:\phi\:\left({argument}\:{of}\:\mathrm{1}−{z}^{\mathrm{4}} \right)=\:\mathrm{2}\theta−\frac{\pi}{\mathrm{2}}\:. \\ $$ | ||
Commented byrahul 19 last updated on 10/Sep/18 | ||
$$\mathrm{thanks}\:\mathrm{sir}. \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18 | ||
$${z}={cos}\theta+{isin}\theta={e}^{{i}\theta} \\ $$ $$\mathrm{1}−{z}^{\mathrm{4}} \\ $$ $$=\mathrm{1}−{e}^{{i}\mathrm{4}\theta} \\ $$ $$=\mathrm{1}−\left({cos}\mathrm{4}\theta+{isin}\mathrm{4}\theta\right) \\ $$ $$=\mathrm{1}−{cos}\mathrm{4}\theta−{isin}\mathrm{4}\theta \\ $$ $$=\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta−{i}×\mathrm{2}{sin}\mathrm{2}\theta.{cos}\mathrm{2}\theta \\ $$ $${so}\:{tan}\alpha=\frac{−\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta}{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta}=−{cot}\mathrm{2}\theta \\ $$ $${tan}\propto=−{tan}\left(\frac{\Pi}{\mathrm{2}}−\mathrm{2}\theta\right) \\ $$ $${tan}\alpha={tan}\left(\mathrm{2}\theta−\frac{\Pi}{\mathrm{2}}\right) \\ $$ $$\alpha=\mathrm{2}\theta−\frac{\Pi}{\mathrm{2}} \\ $$ | ||
Commented byrahul 19 last updated on 10/Sep/18 | ||
$$\mathrm{thanks}\:\mathrm{sir}. \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18 | ||
$${z}={cos}\theta+{isin}\theta={e}^{{i}\theta} \\ $$ $$\mathrm{1}−{z}^{\mathrm{4}} \\ $$ $$=\mathrm{1}−{e}^{{i}\mathrm{4}\theta} \\ $$ $$=\mathrm{1}−\left({cos}\mathrm{4}\theta+{isin}\mathrm{4}\theta\right) \\ $$ $$=\mathrm{1}−{cos}\mathrm{4}\theta−{isin}\mathrm{4}\theta \\ $$ $$=\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta−{i}×\mathrm{2}{sin}\mathrm{2}\theta.{cos}\mathrm{2}\theta \\ $$ $${so}\:{tan}\alpha=\frac{−\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta}{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta}=−{cot}\mathrm{2}\theta \\ $$ $${tan}\propto=−{tan}\left(\frac{\Pi}{\mathrm{2}}−\mathrm{2}\theta\right) \\ $$ $${tan}\alpha={tan}\left(\mathrm{2}\theta−\frac{\Pi}{\mathrm{2}}\right) \\ $$ $$\alpha=\mathrm{2}\theta−\frac{\Pi}{\mathrm{2}} \\ $$ | ||