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Question Number 168118 by Mastermind last updated on 03/Apr/22

If tan(θ+iφ)=cosα+isinα, prove that : θ=((nΠ)/2)+(Π/4) and φ=(1/2)log tan((Π/4)+(α/2))

$${If}\:{tan}\left(\theta+{i}\phi\right)={cos}\alpha+{isin}\alpha,\: \\ $$$${prove}\:{that}\::\:\theta=\frac{{n}\Pi}{\mathrm{2}}+\frac{\Pi}{\mathrm{4}}\:{and}\:\phi=\frac{\mathrm{1}}{\mathrm{2}}{log}\:{tan}\left(\frac{\Pi}{\mathrm{4}}+\frac{\alpha}{\mathrm{2}}\right) \\ $$

Answered by mathsmine last updated on 05/Apr/22

tan(θ+i∅)=((tan(θ)+tan(i∅))/(1−tan(θ)tan(i∅))) tan(ix)=ith(x) ⇔((tan(θ)+ith(∅))/(1−itan(θ)th(∅)))=cos(a)+isin(a) ⇔cos(a)+sin(a)tan(θ)th(∅)+i(sin(a)−tan(θ)th(∅)cos(a)) =tan(θ)+ith(∅) tan(θ)=((cos(a))/(1−sin(a)th(∅))) th(∅)=((sin(a))/(1+tg(θ)cos(a))) ⇒tg(θ)=cos(a)(1+tg(θ)cos(a)).(1/((1+tg(θ)cos(a)−sin^2 (a))) =((1+tg(θ)cos(a))/(tg(θ)+cos(a)))⇔ tg^2 (θ)=1⇒ tg(θ)=+_− 1⇒ θ=(π/4)+n(π/2) th(∅)=((sin(a))/(1+_− cos(a))), ((sin(a))/(1+cos(a)))=tg((a/2)).....E ((sin(a))/(1−cos(a)))=cot((a/2)) th(x)=a⇒x=argth(a)=(1/2)ln(((a+1)/(1−a))) E⇒∅=(1/2)ln(((1+tg((a/2)))/(1−tg((a/2))))))=(1/2)ln(tg((π/4)+(a/2))) ≤((1+tg(x))/(1−tg(x)))=tg(x+(π/4))≥ So θ=((nπ)/2)+(π/4) ∅=(1/2)ln(tg((a/2)+(π/4)))

$${tan}\left(\theta+{i}\emptyset\right)=\frac{{tan}\left(\theta\right)+{tan}\left({i}\emptyset\right)}{\mathrm{1}−{tan}\left(\theta\right){tan}\left({i}\emptyset\right)} \\ $$$${tan}\left({ix}\right)={ith}\left({x}\right) \\ $$$$\Leftrightarrow\frac{{tan}\left(\theta\right)+{ith}\left(\emptyset\right)}{\mathrm{1}−{itan}\left(\theta\right){th}\left(\emptyset\right)}={cos}\left({a}\right)+{isin}\left({a}\right) \\ $$$$\Leftrightarrow{cos}\left({a}\right)+{sin}\left({a}\right){tan}\left(\theta\right){th}\left(\emptyset\right)+{i}\left({sin}\left({a}\right)−{tan}\left(\theta\right){th}\left(\emptyset\right){cos}\left({a}\right)\right) \\ $$$$={tan}\left(\theta\right)+{ith}\left(\emptyset\right) \\ $$$${tan}\left(\theta\right)=\frac{{cos}\left({a}\right)}{\mathrm{1}−{sin}\left({a}\right){th}\left(\emptyset\right)} \\ $$$${th}\left(\emptyset\right)=\frac{{sin}\left({a}\right)}{\mathrm{1}+{tg}\left(\theta\right){cos}\left({a}\right)} \\ $$$$\Rightarrow{tg}\left(\theta\right)={cos}\left({a}\right)\left(\mathrm{1}+{tg}\left(\theta\right){cos}\left({a}\right)\right).\frac{\mathrm{1}}{\left(\mathrm{1}+{tg}\left(\theta\right){cos}\left({a}\right)−{sin}^{\mathrm{2}} \left({a}\right)\right.} \\ $$$$=\frac{\mathrm{1}+{tg}\left(\theta\right){cos}\left({a}\right)}{{tg}\left(\theta\right)+{cos}\left({a}\right)}\Leftrightarrow \\ $$$${tg}^{\mathrm{2}} \left(\theta\right)=\mathrm{1}\Rightarrow\:{tg}\left(\theta\right)=\underset{−} {+}\mathrm{1}\Rightarrow \\ $$$$\theta=\frac{\pi}{\mathrm{4}}+{n}\frac{\pi}{\mathrm{2}} \\ $$$${th}\left(\emptyset\right)=\frac{{sin}\left({a}\right)}{\mathrm{1}\underset{−} {+}{cos}\left({a}\right)}, \\ $$$$\frac{{sin}\left({a}\right)}{\mathrm{1}+{cos}\left({a}\right)}={tg}\left(\frac{{a}}{\mathrm{2}}\right).....{E} \\ $$$$\frac{{sin}\left({a}\right)}{\mathrm{1}−{cos}\left({a}\right)}={cot}\left(\frac{{a}}{\mathrm{2}}\right) \\ $$$${th}\left({x}\right)={a}\Rightarrow{x}={argth}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{a}+\mathrm{1}}{\mathrm{1}−{a}}\right) \\ $$$${E}\Rightarrow\emptyset=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{tg}\left(\frac{{a}}{\mathrm{2}}\right)}{\left.\mathrm{1}−{tg}\left(\frac{{a}}{\mathrm{2}}\right)\right)}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}+\frac{{a}}{\mathrm{2}}\right)\right) \\ $$$$\leqslant\frac{\mathrm{1}+{tg}\left({x}\right)}{\mathrm{1}−{tg}\left({x}\right)}={tg}\left({x}+\frac{\pi}{\mathrm{4}}\right)\geqslant \\ $$$${So}\:\theta=\frac{{n}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}} \\ $$$$\emptyset=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({tg}\left(\frac{{a}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Mastermind last updated on 05/Apr/22

Thanks man, you did a great Job! Could you please drop your Whatsapp number?

$${Thanks}\:{man},\:{you}\:{did}\:{a}\:{great}\:{Job}! \\ $$$$ \\ $$$${Could}\:{you}\:{please}\:{drop}\:{your}\:{Whatsapp}\:{number}? \\ $$

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