Question Number 145524 by physicstutes last updated on 05/Jul/21
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{argument}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{below} \\ $$$$\left(\mathrm{i}\right)\:{z}\:=\:\mathrm{1}+{e}^{\frac{\pi}{\mathrm{6}}{i}} \\ $$$$\left(\mathrm{ii}\right)\:{z}\:=\:\mathrm{1}\:−{e}^{\frac{\pi}{\mathrm{6}}{i}} \\ $$
Answered by Olaf_Thorendsen last updated on 05/Jul/21
![(i) z = 1+e^(i(π/6)) z = e^(i(π/(12))) (e^(−i(π/(12))) +e^(i(π/(12))) ) z = 2cos(π/(12))e^(i(π/(12))) argz = (π/(12)) [2π] (ii) z = 1 −e^((π/6)i) z = e^(i(π/(12))) (e^(−i(π/(12))) −e^(i(π/(12))) ) z = −2isin(π/(12))e^(i(π/(12))) z = e^(iπ) 2e^(i(π/2)) sin(π/(12))e^(i(π/(12))) z = 2sin(π/(12))e^(i(π+(π/2)+(π/(12)))) z = 2sin(π/(12))e^(i((19π)/(12))) argz = ((19π)/(12)) [2π]](https://www.tinkutara.com/question/Q145526.png)
$$\left({i}\right)\:{z}\:=\:\mathrm{1}+{e}^{{i}\frac{\pi}{\mathrm{6}}} \\ $$$${z}\:=\:{e}^{{i}\frac{\pi}{\mathrm{12}}} \left({e}^{−{i}\frac{\pi}{\mathrm{12}}} +{e}^{{i}\frac{\pi}{\mathrm{12}}} \right) \\ $$$${z}\:=\:\mathrm{2cos}\frac{\pi}{\mathrm{12}}{e}^{{i}\frac{\pi}{\mathrm{12}}} \\ $$$$\mathrm{arg}{z}\:=\:\frac{\pi}{\mathrm{12}}\:\left[\mathrm{2}\pi\right] \\ $$$$\left(\mathrm{ii}\right)\:{z}\:=\:\mathrm{1}\:−{e}^{\frac{\pi}{\mathrm{6}}{i}} \\ $$$${z}\:=\:{e}^{{i}\frac{\pi}{\mathrm{12}}} \left({e}^{−{i}\frac{\pi}{\mathrm{12}}} \:−{e}^{{i}\frac{\pi}{\mathrm{12}}} \right) \\ $$$${z}\:=\:−\mathrm{2}{i}\mathrm{sin}\frac{\pi}{\mathrm{12}}{e}^{{i}\frac{\pi}{\mathrm{12}}} \\ $$$${z}\:=\:{e}^{{i}\pi} \mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{2}}} \mathrm{sin}\frac{\pi}{\mathrm{12}}{e}^{{i}\frac{\pi}{\mathrm{12}}} \\ $$$${z}\:=\:\mathrm{2sin}\frac{\pi}{\mathrm{12}}{e}^{{i}\left(\pi+\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{12}}\right)} \\ $$$${z}\:=\:\mathrm{2sin}\frac{\pi}{\mathrm{12}}{e}^{{i}\frac{\mathrm{19}\pi}{\mathrm{12}}} \\ $$$$\mathrm{arg}{z}\:=\:\frac{\mathrm{19}\pi}{\mathrm{12}}\:\left[\mathrm{2}\pi\right] \\ $$
Commented by physicstutes last updated on 05/Jul/21
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$ \\ $$