# lets-u-v-w-C-3-such-that-uvw-0-proof-that-or-give-a-counter-example-v-u-w-v-u-v-w-u-w-u-v-w-0-

Question Number 3410 by 123456 last updated on 13/Dec/15
$$\mathrm{lets}\:\left({u},{v},{w}\right)\in\mathbb{C}^{\mathrm{3}} \:\mathrm{such}\:\mathrm{that}\:{uvw}\neq\mathrm{0},\:\mathrm{proof} \\$$$$\mathrm{that}\:\left(\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\right) \\$$$$\frac{{v}}{{u}}=\frac{{w}}{{v}}\wedge\mid{u}\mid=\mid{v}\mid=\mid{w}\mid\wedge{u}\neq{w}\Rightarrow{u}+{v}+{w}=\mathrm{0} \\$$
Commented by RasheedSindhi last updated on 13/Dec/15
$$\mathcal{W}{hat}\:{is}\:{meant}\:{by}\:{u},{v},{w}\in\mathbb{C}^{\mathrm{3}} \:? \\$$
Commented by 123456 last updated on 13/Dec/15
$${u}\in\mathbb{C}\wedge{v}\in\mathbb{C}\wedge{w}\in\mathbb{C} \\$$
Commented by RasheedSindhi last updated on 13/Dec/15
$${Do}\:{you}\:{mean} \\$$$${Given}:\left({u},{v},{w}\right)\in\mathbb{C}^{\mathrm{3}} \:\wedge\:{uvw}\neq\mathrm{0} \\$$$$\:\:{To}\:{prove}:\frac{{v}}{{u}}=\frac{{w}}{{v}}\wedge\mid{u}\mid=\mid{v}\mid=\mid{w}\mid\wedge{u}\neq{w}\Rightarrow{u}+{v}+{w}=\mathrm{0}? \\$$$$\\$$
Commented by 123456 last updated on 13/Dec/15
$$\mathrm{yes}\:\left(\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{conter}\:\mathrm{example}\right) \\$$
Commented by Yozzi last updated on 13/Dec/15
$${v}^{\mathrm{2}} ={uw}.\:{v}=\mid{v}\mid{e}^{{i}\theta} ,\:{u}=\mid{u}\mid{e}^{{i}\alpha} ,\:{w}=\mid{w}\mid{e}^{{i}\beta} \\$$$$\mid{v}\mid^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} =\mid{u}\mid\mid{w}\mid{e}^{{i}\left(\alpha+\beta\right)} =\mid{v}\mid^{\mathrm{2}} {e}^{{i}\left(\alpha+\beta\right)} \\$$$${e}^{\mathrm{2}{i}\theta} ={e}^{{i}\left(\alpha+\beta\right)} \\$$$$\Rightarrow{cos}\mathrm{2}\theta={cos}\left(\alpha+\beta\right) \\$$$${sin}\mathrm{2}\theta={sin}\left(\alpha+\beta\right) \\$$$${u}\neq{w}\Rightarrow\alpha\neq\beta.\: \\$$$${sin}\mathrm{2}\theta={sin}\left(\alpha+\beta\right): \\$$$$\mathrm{2}{cos}\frac{\mathrm{2}\theta+\alpha+\beta}{\mathrm{2}}{sin}\frac{\mathrm{2}\theta−\alpha−\beta}{\mathrm{2}}=\mathrm{0} \\$$$$\Rightarrow\mathrm{2}\theta−\alpha−\beta=\mathrm{2}{n}\pi\Rightarrow\theta=\frac{\alpha+\beta}{\mathrm{2}}+{n}\pi\:\:\:{n}\in\mathbb{Z} \\$$$$\frac{\mathrm{2}\theta+\alpha+\beta}{\mathrm{2}}=\mathrm{2}{r}\pi\pm\mathrm{0}.\mathrm{5}\pi \\$$$$\mathrm{2}\theta+\alpha+\beta=\left(\mathrm{4}{r}\pm\mathrm{1}\right)\pi \\$$$$\theta=\frac{\left(\mathrm{4}{r}\pm\mathrm{1}\right)\pi−\alpha−\beta}{\mathrm{2}}\:\:\:\:\:{r}\in\mathbb{Z} \\$$$$\:\:\:−\pi<\theta\leqslant\pi\:. \\$$$$\frac{{v}}{{u}}=\frac{{w}}{{v}}\Rightarrow{e}^{{i}\left(\theta−\alpha\right)} ={e}^{{i}\left(\beta−\theta\right)} \\$$$$\Rightarrow\:\:\:{cos}\left(\theta−\alpha\right)={cos}\left(\beta−\theta\right) \\$$$${sin}\left(\theta−\alpha\right)={sin}\left(\beta−\theta\right) \\$$$${let}\:\theta=\mathrm{0},\:\alpha=\pi/\mathrm{2},\:\beta=−\pi/\mathrm{2}\:\left(\alpha\neq\beta\right) \\$$$$\therefore\:{v}+{w}+{u}=\mid{v}\mid+\mid{w}\mid\left(\mathrm{0}−{i}\right)+\mid{u}\mid\left(\mathrm{0}+{i}\right) \\$$$$=\mid{v}\mid+{i}\left(−\mid{w}\mid+\mid{u}\mid\right) \\$$$$=\mid{v}\mid\neq\mathrm{0}\:{if}\:\frac{{v}}{{u}}=\frac{{w}}{{v}}\:{and}\:\mid{v}\mid=\mid{w}\mid=\mid{u}\mid. \\$$$${uvw}=\mid{v}\mid^{\mathrm{3}} ×\left(−{i}\right){i}=\mid{v}\mid^{\mathrm{3}} \neq\mathrm{0} \\$$$$\mathbb{R}\subset\mathbb{C}\:{so}\:{v}=\mid{v}\mid\in\mathbb{C}\:,\:{u}=\mid{u}\mid{i}\in\mathbb{C},\:{w}=−\mid{w}\mid{i}\in\mathbb{C} \\$$$${We}\:{can}\:{set}\:\mid{v}\mid=\mid{w}\mid=\mid{u}\mid={r}\:{for}\:{r}>\mathrm{0},\:{r}\in\mathbb{R}. \\$$$$\\$$
Commented by Yozzii last updated on 13/Dec/15
$${Testing}\:{new}\:{feature} \\$$
Answered by prakash jain last updated on 13/Dec/15
$$\mathrm{Counter}\:\mathrm{Example}\:\mathrm{from}\:\mathrm{Yozzi}'\mathrm{s}\:\mathrm{comment} \\$$$${a},−{ai},{ai} \\$$