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Question Number 197409 by mathlove last updated on 16/Sep/23

Commented by mathlove last updated on 16/Sep/23

$$anysolutioniscorectand$$$$anyoneisancorect$$

Commented by MM42 last updated on 16/Sep/23

$$1=\sqrt{1}=\sqrt{-1\times -1}=\sqrt{-1}\times \sqrt{-1}=i\times i=-1$$

Commented by mahdipoor last updated on 16/Sep/23

$$\sqrt{e^{2a+bi}}\times \sqrt{e^{2x+yi}}\stackrel{?}{=}\sqrt{e^{2a+bi}\times e^{2x+yi}}$$$$e^{a+b_{n}i}\times e^{x+y_{n}i}\stackrel{?}{=}\sqrt{e^{2(a+x)+(b+y)i}}$$$$e^{(a+x)+(b_n+y_n)i}\stackrel{?}{=}{e}^{(a+x)+(b_n+y_n)i}$$$$\Rightarrow b_n+y_{n}mustsameforLandR$$$${\phi }_n=\frac{\phi }{2}and\frac{\phi }{2}+\pi$$$$\sqrt{-1}\times \sqrt{-1}=\sqrt{-1\times -1}$$$$\Rightarrow \sqrt{e^{\pi i}}\times \sqrt{e^{\pi i}}=\sqrt{e^{2\pi i}}$$$$\Rightarrow e^{\frac{\pi }{2}i}\times e^{\frac{\pi }{2}i}=e^{(\pi +\pi )i}$$$$\Rightarrow -1=1$$

Commented by Alleddawi last updated on 17/Sep/23

$$Thesecondmathematicalexpressioniscorrect.$$$$Because\sqrt{a}\times \sqrt{b}=\sqrt{a\times b},ifatmostoneofaandbisnegative.$$

Answered by Frix last updated on 16/Sep/23

$$x_j=r_j{\mathrm{e}}^{\mathrm{i}{\theta }_j};r_j⩾0\wedge -\pi <{\theta }_j⩽\pi$$$$x_3=x_1{x}_2=r_1{\mathrm{e}}^{\mathrm{i}{\theta }_1}{r}_2{\mathrm{e}}^{\mathrm{i}{\theta }_2}=r_1{r}_2{\mathrm{e}}^{\mathrm{i}({\theta }_1+{\theta }_2)}=r_3{\mathrm{e}}^{\mathrm{i}{\overline{\theta }}_3}$$$$x_4=x^k={\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}^k=r^k{\mathrm{e}}^{\mathrm{i}k\theta }=r_4{\mathrm{e}}^{\mathrm{i}{\overline{\theta }}_4}$$$$\mathrm{We}\mathrm{must}\mathrm{have}-\pi <{\theta }_j⩽\pi \Rightarrow$$$${\theta }_j=\mathrm{mod}({\overline{\theta }}_j,\pi )-\frac{\pi }{2}(1-\mathrm{sign}\sin {\overline{\theta }}_j)$$$$[{\overline{\theta }}_j=2n\pi \Rightarrow {\theta }_j=0]$$$$$$$$\sqrt{-4}\times \sqrt{-9}=\sqrt{{4\mathrm{e}}^{\mathrm{i}\pi }}\times \sqrt{{9\mathrm{e}}^{\mathrm{i}\pi }}=\sqrt{4}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\sqrt{9}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}=$$$$=2\times 3\times {\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}+\mathrm{i}\frac{\pi }{2}}={6\mathrm{e}}^{\mathrm{i}\pi }=-6$$$$\sqrt{(-4)(-9)}=\sqrt{{4\mathrm{e}}^{\mathrm{i}\pi }\times {9\mathrm{e}}^{\mathrm{i}\pi }}=\sqrt{{36\mathrm{e}}^{2\mathrm{i}\pi }}=\sqrt{{36\mathrm{e}}^0}=$$$$=\sqrt{36}=6$$$$$$$$\mathrm{All}\mathrm{depends}\mathrm{on}\mathrm{the}\mathrm{correct}\mathrm{execution}\mathrm{of}\mathrm{the}$$$$\mathrm{rules}.\mathrm{If}\mathrm{you}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{rules},\mathrm{learn}\mathrm{them}.$$$$\mathrm{Sadly}\mathrm{on}\mathrm{the}\mathrm{www}\mathrm{there}\mathrm{is}\mathrm{more}\mathrm{opinion}\mathrm{than}$$$$\mathrm{knowledge}.\mathrm{The}\mathrm{set}\mathrm{of}\mathrm{rules}\mathrm{was}\mathrm{made}\mathrm{to}\mathrm{get}$$$$\mathrm{consistency}.$$

Commented by Frix last updated on 16/Sep/23

$$\mathrm{Btw}\mathrm{the}\mathrm{reason}\mathrm{for}-\pi <\theta ⩽\pi \mathrm{is}\mathrm{to}\mathrm{get}\mathrm{useful}$$$$\mathrm{results}\mathrm{for}\mathrm{conjugated}\mathrm{numbers}$$$$\sqrt{2+2\sqrt{3}\mathrm{i}}=\sqrt{3}+\mathrm{i}$$$$\sqrt{2-2\sqrt{3}\mathrm{i}}=\sqrt{3}-\mathrm{i}$$$$2+2\sqrt{3}\mathrm{i}={4\mathrm{e}}^{\mathrm{i}\frac{\pi }{3}};\sqrt{{4\mathrm{e}}^{\mathrm{i}\frac{\pi }{3}}}={2\mathrm{e}}^{\mathrm{i}\frac{\pi }{6}}=\sqrt{3}+\mathrm{i}$$$$2-2\sqrt{3}\mathrm{i}={4\mathrm{e}}^{-\mathrm{i}\frac{\pi }{3}};\sqrt{{4\mathrm{e}}^{-\mathrm{i}\frac{\pi }{3}}}={2\mathrm{e}}^{-\mathrm{i}\frac{\pi }{6}}=\sqrt{3}-\mathrm{i}$$$$\mathrm{If}{\mathrm{we}}^{\prime }\mathrm{d}\mathrm{use}0⩽\theta <2\pi$$$$2-2\sqrt{3}\mathrm{i}={4\mathrm{e}}^{\mathrm{i}\frac{5\pi }{3}};\sqrt{{4\mathrm{e}}^{\mathrm{i}\frac{5\pi }{3}}}={2\mathrm{e}}^{\mathrm{i}\frac{5\pi }{6}}=-\sqrt{3}+\mathrm{i}$$

Commented by mathlove last updated on 17/Sep/23

$$thanksforall$$

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