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Let-z-1-cos-10pi-9-isin-10pi-9-Then-A-z-2cos-2pi-9-B-arg-z-8pi-9-C-z-2cos-4pi-9-D-arg-z-5pi-9-

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Question Number 140104 by EnterUsername last updated on 04/May/21
Let z=1+cos(10π/9)+isin(10π/9). Then (A) ∣z∣=2cos(((2π)/9)) (B) arg z=((8π)/9) (C) ∣z∣=2cos(((4π)/9)) (D) arg z=((5π)/9)
$$\mathrm{Let}\:{z}=\mathrm{1}+\mathrm{cos}\left(\mathrm{10}\pi/\mathrm{9}\right)+{i}\mathrm{sin}\left(\mathrm{10}\pi/\mathrm{9}\right).\:\mathrm{Then} \\ $$$$\left(\mathrm{A}\right)\:\mid{z}\mid=\mathrm{2cos}\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{arg}\:{z}=\frac{\mathrm{8}\pi}{\mathrm{9}} \\ $$$$\left(\mathrm{C}\right)\:\mid{z}\mid=\mathrm{2cos}\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{arg}\:{z}=\frac{\mathrm{5}\pi}{\mathrm{9}} \\ $$
Commented by EnterUsername last updated on 04/May/21
One or more answers may be correct.
$$\mathrm{One}\:\mathrm{or}\:\mathrm{more}\:\mathrm{answers}\:\mathrm{may}\:\mathrm{be}\:\mathrm{correct}. \\ $$
Answered by MJS_new last updated on 04/May/21
cos ((10π)/9) +i sin ((10π)/9) =−cos (π/9) −i sin (π/9) ∣z∣=(√((1−cos θ)^2 +(−sin θ)^2 ))=(√(2−2cos θ))= =2∣sin (θ/2)∣=2sin (π/(18)) =2cos ((4π)/9) tan (arg z)=((−sin θ)/(1−cos θ))=−cot (θ/2) =−cot (π/(18)) ⇒ ⇒ arg z =((5π)/9)
$$\mathrm{cos}\:\frac{\mathrm{10}\pi}{\mathrm{9}}\:+\mathrm{i}\:\mathrm{sin}\:\frac{\mathrm{10}\pi}{\mathrm{9}}\:=−\mathrm{cos}\:\frac{\pi}{\mathrm{9}}\:−\mathrm{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{9}} \\ $$$$\mid{z}\mid=\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left(−\mathrm{sin}\:\theta\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}−\mathrm{2cos}\:\theta}= \\ $$$$=\mathrm{2}\mid\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\mid=\mathrm{2sin}\:\frac{\pi}{\mathrm{18}}\:=\mathrm{2cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}} \\ $$$$\mathrm{tan}\:\left(\mathrm{arg}\:{z}\right)=\frac{−\mathrm{sin}\:\theta}{\mathrm{1}−\mathrm{cos}\:\theta}=−\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\:=−\mathrm{cot}\:\frac{\pi}{\mathrm{18}}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{arg}\:{z}\:=\frac{\mathrm{5}\pi}{\mathrm{9}} \\ $$
Commented by EnterUsername last updated on 04/May/21
Thank you Sir arg z=−tan^(−1) (cot(π/(18)))=(π/2)+cot^(−1) (cot(π/(18)))=(π/2)+(π/(18))=((5π)/9) Understood !
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$$$\mathrm{arg}\:{z}=−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}\frac{\pi}{\mathrm{18}}\right)=\frac{\pi}{\mathrm{2}}+\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{cot}\frac{\pi}{\mathrm{18}}\right)=\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{18}}=\frac{\mathrm{5}\pi}{\mathrm{9}} \\ $$$$\mathrm{Understood}\:! \\ $$

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