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Question Number 109147 by ajfour last updated on 21/Aug/20

The principal argument of z=1+cos (((6π)/5))+isin (((6π)/5)) is = ?

$${The}\:{principal}\:{argument}\:{of} \\ $$$${z}=\mathrm{1}+\mathrm{cos}\:\left(\frac{\mathrm{6}\pi}{\mathrm{5}}\right)+{i}\mathrm{sin}\:\left(\frac{\mathrm{6}\pi}{\mathrm{5}}\right)\:\:\:{is}\:=\:? \\ $$

Answered by Dwaipayan Shikari last updated on 21/Aug/20

tanθ=((sin((6π)/5))/(1+cos((6π)/5)))=tan((3π)/5) θ=kπ+((3π)/5) arg(z)=kπ+((3π)/5) (k∈Z) arg(z)=−π+((3π)/5)=−((2π)/5)

$$ \\ $$$${tan}\theta=\frac{{sin}\frac{\mathrm{6}\pi}{\mathrm{5}}}{\mathrm{1}+{cos}\frac{\mathrm{6}\pi}{\mathrm{5}}}={tan}\frac{\mathrm{3}\pi}{\mathrm{5}} \\ $$$$\theta={k}\pi+\frac{\mathrm{3}\pi}{\mathrm{5}} \\ $$$${arg}\left({z}\right)={k}\pi+\frac{\mathrm{3}\pi}{\mathrm{5}}\:\:\left({k}\in\mathbb{Z}\right) \\ $$$${arg}\left({z}\right)=−\pi+\frac{\mathrm{3}\pi}{\mathrm{5}}=−\frac{\mathrm{2}\pi}{\mathrm{5}} \\ $$

Commented by ajfour last updated on 21/Aug/20

i think answer should be θ=−((2π)/5).

$${i}\:{think}\:\:{answer}\:{should}\:{be}\:\theta=−\frac{\mathrm{2}\pi}{\mathrm{5}}. \\ $$

Commented by Dwaipayan Shikari last updated on 21/Aug/20

Yes. I have edited my mistake

$${Yes}.\:{I}\:{have}\:{edited}\:{my}\:{mistake} \\ $$

Answered by Dwaipayan Shikari last updated on 21/Aug/20

z=1−((((√5)−1)/4))−i(((√(10−2(√5)))/4)) z=((−(√5)+5)/4)−i(((√(10−2(√5)))/4)) z=((5−(√5))/4)−i(((√(10−2(√5)))/4)) arg(z)=tan^(−1) (−((√(10−2(√5)))/(5−(√5))))=−((2π)/5)

$${z}=\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)−{i}\left(\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\right) \\ $$$${z}=\frac{−\sqrt{\mathrm{5}}+\mathrm{5}}{\mathrm{4}}−{i}\left(\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\right) \\ $$$${z}=\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{4}}−{i}\left(\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\right) \\ $$$${arg}\left({z}\right)={tan}^{−\mathrm{1}} \left(−\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{5}−\sqrt{\mathrm{5}}}\right)=−\frac{\mathrm{2}\pi}{\mathrm{5}} \\ $$

Answered by pticantor last updated on 21/Aug/20

z=1+e^(iθ) =e^(i(θ/2)) (e^(−i(θ/2)) +e^(i(θ/2)) ) =2cos((θ/2))e^(i(θ/2)) take θ=((6π)/5) and arg(z)=(θ/2)=((3π)/5)

$${z}=\mathrm{1}+{e}^{{i}\theta} ={e}^{{i}\frac{\theta}{\mathrm{2}}} \left({e}^{−{i}\frac{\theta}{\mathrm{2}}} +{e}^{{i}\frac{\theta}{\mathrm{2}}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{cos}\left(\frac{\theta}{\mathrm{2}}\right){e}^{{i}\frac{\theta}{\mathrm{2}}} \\ $$$${take}\:\theta=\frac{\mathrm{6}\pi}{\mathrm{5}}\:{and}\:{arg}\left({z}\right)=\frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}\pi}{\mathrm{5}} \\ $$

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