Question Number 197409 by mathlove last updated on 16/Sep/23
Commented by mathlove last updated on 16/Sep/23
$${any}\:{solution}\:{is}\:{corect}\:{and}\: \\ $$$${any}\:{one}\:{is}\:{ancorect}\: \\ $$
Commented by AST last updated on 16/Sep/23
$$\sqrt{{a}×{b}}=\sqrt{{a}}×\sqrt{{b}}\:{is}\:{only}\:{true}\:{if}\:{both}\:{a}\:{and}\:{b}\:{are} \\ $$$${not}\:{negative} \\ $$
Commented by MM42 last updated on 16/Sep/23
$$\mathrm{1}=\sqrt{\mathrm{1}}=\sqrt{−\mathrm{1}×−\mathrm{1}}=\sqrt{−\mathrm{1}}×\sqrt{−\mathrm{1}}={i}×{i}=−\mathrm{1} \\ $$
Commented by mahdipoor last updated on 16/Sep/23
$$\sqrt{{e}^{\mathrm{2}{a}+{bi}} }×\sqrt{{e}^{\mathrm{2}{x}+{yi}} }\overset{?} {=}\sqrt{{e}^{\mathrm{2}{a}+{bi}} ×{e}^{\mathrm{2}{x}+{yi}} } \\ $$$${e}^{{a}+{b}_{{n}} {i}} ×{e}^{{x}+{y}_{{n}} {i}} \overset{?} {=}\sqrt{{e}^{\mathrm{2}\left({a}+{x}\right)+\left({b}+{y}\right){i}} } \\ $$$${e}^{\left({a}+{x}\right)+\left({b}_{{n}} +{y}_{{n}} \right){i}} \overset{?} {=}{e}^{\left({a}+{x}\right)+\left({b}_{{n}} +{y}_{{n}} \right){i}} \\ $$$$\Rightarrow{b}_{{n}} +{y}_{{n}} \:\:{must}\:{same}\:{for}\:{L}\:{and}\:{R} \\ $$$$\varphi_{{n}} =\frac{\varphi}{\mathrm{2}}\:{and}\:\:\frac{\varphi}{\mathrm{2}}+\pi \\ $$$$\sqrt{−\mathrm{1}}×\sqrt{−\mathrm{1}}=\sqrt{−\mathrm{1}×−\mathrm{1}} \\ $$$$\Rightarrow\sqrt{{e}^{\pi{i}} }×\sqrt{{e}^{\pi{i}} }=\sqrt{{e}^{\mathrm{2}\pi{i}} } \\ $$$$\Rightarrow{e}^{\frac{\pi}{\mathrm{2}}{i}} ×{e}^{\frac{\pi}{\mathrm{2}}{i}} ={e}^{\left(\pi+\pi\right){i}\:\:\:\:\:\:\:\:\:\:} \\ $$$$\Rightarrow−\mathrm{1}=\mathrm{1} \\ $$
Commented by Alleddawi last updated on 17/Sep/23
$${The}\:{second}\:{mathematical}\:{expression}\:{is}\:{correct}. \\ $$$${Because}\:\sqrt{{a}}×\sqrt{{b}}=\sqrt{{a}×{b}\:}\:\:,\:{if}\:{at}\:{most}\:{one}\:{of}\:{a}\:{and}\:{b}\:{is}\:{negative}. \\ $$
Answered by Frix last updated on 16/Sep/23
![x_j =r_j e^(iθ_j ) ; r_j ≥0∧−π<θ_j ≤π x_3 =x_1 x_2 =r_1 e^(iθ_1 ) r_2 e^(iθ_2 ) =r_1 r_2 e^(i(θ_1 +θ_2 )) =r_3 e^(iθ_3 ^ ) x_4 =x^k =(re^(iθ) )^k =r^k e^(ikθ) =r_4 e^(iθ_4 ^ ) We must have −π<θ_j ≤π ⇒ θ_j =mod (θ_j ^ , π) −(π/2)(1−sign sin θ_j ^ ) [θ_j ^ =2nπ ⇒ θ_j =0] (√(−4))×(√(−9))=(√(4e^(iπ) ))×(√(9e^(iπ) ))=(√4)e^(i(π/2)) (√9)e^(i(π/2)) = =2×3×e^(i(π/2)+i(π/2)) =6e^(iπ) =−6 (√((−4)(−9)))=(√(4e^(iπ) ×9e^(iπ) ))=(√(36e^(2iπ) ))=(√(36e^0 ))= =(√(36))=6 All depends on the correct execution of the rules. If you don′t know the rules, learn them. Sadly on the www there is more opinion than knowledge. The set of rules was made to get consistency.](https://www.tinkutara.com/question/Q197420.png)
$${x}_{{j}} ={r}_{{j}} \mathrm{e}^{\mathrm{i}\theta_{{j}} } ;\:{r}_{{j}} \geqslant\mathrm{0}\wedge−\pi<\theta_{{j}} \leqslant\pi \\ $$$${x}_{\mathrm{3}} ={x}_{\mathrm{1}} {x}_{\mathrm{2}} ={r}_{\mathrm{1}} \mathrm{e}^{\mathrm{i}\theta_{\mathrm{1}} } {r}_{\mathrm{2}} \mathrm{e}^{\mathrm{i}\theta_{\mathrm{2}} } ={r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{e}^{\mathrm{i}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} ={r}_{\mathrm{3}} \mathrm{e}^{\mathrm{i}\bar {\theta}_{\mathrm{3}} } \\ $$$${x}_{\mathrm{4}} ={x}^{{k}} =\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{k}} ={r}^{{k}} \mathrm{e}^{\mathrm{i}{k}\theta} ={r}_{\mathrm{4}} \mathrm{e}^{\mathrm{i}\bar {\theta}_{\mathrm{4}} } \\ $$$$\mathrm{We}\:\mathrm{must}\:\mathrm{have}\:−\pi<\theta_{{j}} \leqslant\pi\:\Rightarrow \\ $$$$\theta_{{j}} =\mathrm{mod}\:\left(\bar {\theta}_{{j}} ,\:\pi\right)\:−\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{sign}\:\mathrm{sin}\:\bar {\theta}_{{j}} \right) \\ $$$$\:\:\:\:\:\left[\bar {\theta}_{{j}} =\mathrm{2}{n}\pi\:\Rightarrow\:\theta_{{j}} =\mathrm{0}\right] \\ $$$$ \\ $$$$\sqrt{−\mathrm{4}}×\sqrt{−\mathrm{9}}=\sqrt{\mathrm{4e}^{\mathrm{i}\pi} }×\sqrt{\mathrm{9e}^{\mathrm{i}\pi} }=\sqrt{\mathrm{4}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{9}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} = \\ $$$$=\mathrm{2}×\mathrm{3}×\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}+\mathrm{i}\frac{\pi}{\mathrm{2}}} =\mathrm{6e}^{\mathrm{i}\pi} =−\mathrm{6} \\ $$$$\sqrt{\left(−\mathrm{4}\right)\left(−\mathrm{9}\right)}=\sqrt{\mathrm{4e}^{\mathrm{i}\pi} ×\mathrm{9e}^{\mathrm{i}\pi} }=\sqrt{\mathrm{36e}^{\mathrm{2i}\pi} }=\sqrt{\mathrm{36e}^{\mathrm{0}} }= \\ $$$$=\sqrt{\mathrm{36}}=\mathrm{6} \\ $$$$ \\ $$$$\mathrm{All}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{execution}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{rules}.\:\mathrm{If}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{rules},\:\mathrm{learn}\:\mathrm{them}. \\ $$$$\mathrm{Sadly}\:\mathrm{on}\:\mathrm{the}\:\mathrm{www}\:\mathrm{there}\:\mathrm{is}\:\mathrm{more}\:\mathrm{opinion}\:\mathrm{than} \\ $$$$\mathrm{knowledge}.\:\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{rules}\:\mathrm{was}\:\mathrm{made}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{consistency}.\: \\ $$
Commented by Frix last updated on 16/Sep/23
$$\mathrm{Btw}\:\mathrm{the}\:\mathrm{reason}\:\mathrm{for}\:−\pi<\theta\leqslant\pi\:\mathrm{is}\:\mathrm{to}\:\mathrm{get}\:\mathrm{useful} \\ $$$$\mathrm{results}\:\mathrm{for}\:\mathrm{conjugated}\:\mathrm{numbers} \\ $$$$\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\mathrm{i}}=\sqrt{\mathrm{3}}+\mathrm{i} \\ $$$$\sqrt{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{i}}=\sqrt{\mathrm{3}}−\mathrm{i} \\ $$$$\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\mathrm{i}=\mathrm{4e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} ;\:\sqrt{\mathrm{4e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} }=\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{6}}} =\sqrt{\mathrm{3}}+\mathrm{i} \\ $$$$\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{i}=\mathrm{4e}^{−\mathrm{i}\frac{\pi}{\mathrm{3}}} ;\:\sqrt{\mathrm{4e}^{−\mathrm{i}\frac{\pi}{\mathrm{3}}} }=\mathrm{2e}^{−\mathrm{i}\frac{\pi}{\mathrm{6}}} =\sqrt{\mathrm{3}}−\mathrm{i} \\ $$$$\mathrm{If}\:\mathrm{we}'\mathrm{d}\:\mathrm{use}\:\mathrm{0}\leqslant\theta<\mathrm{2}\pi \\ $$$$\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{i}=\mathrm{4e}^{\mathrm{i}\frac{\mathrm{5}\pi}{\mathrm{3}}} ;\:\sqrt{\mathrm{4e}^{\mathrm{i}\frac{\mathrm{5}\pi}{\mathrm{3}}} }=\mathrm{2e}^{\mathrm{i}\frac{\mathrm{5}\pi}{\mathrm{6}}} =−\sqrt{\mathrm{3}}+\mathrm{i} \\ $$
Commented by mathlove last updated on 17/Sep/23
$${thanks}\:{for}\:{all} \\ $$