Question Number 49202 by rahul 19 last updated on 04/Dec/18 $$\left.\mathrm{1}\right)\:{If}\:\omega\:{is}\:{an}\:{imaginary}\:{fifth}\:{root}\:{of} \\ $$$${unity},\:{then}\:{find}\:{value}\:{of}\: \\ $$$$\mathrm{log}\:_{\mathrm{2}} \:\mid\mathrm{1}+\omega+\omega^{\mathrm{2}} +\omega^{\mathrm{3}} −\frac{\mathrm{1}}{\omega}\mid\:? \\ $$$$\left.\mathrm{2}\right)\:{Find}\:{value}\:{of}\:: \\ $$$$\left({i}+\sqrt{\mathrm{3}}\right)^{\mathrm{100}} +\left({i}−\sqrt{\mathrm{3}}\right)^{\mathrm{100}} +\mathrm{2}^{\mathrm{100}} \:?…
Question Number 180268 by Noorzai last updated on 09/Nov/22 Answered by Ar Brandon last updated on 09/Nov/22 $$\mathrm{Q121680} \\ $$ Answered by LEKOUMA last updated…
Question Number 49200 by rahul 19 last updated on 04/Dec/18 $$\left.\mathrm{1}\right){Find}\:{the}\:{area}\:{of}\:{the}\:{triangle}\:{formed} \\ $$$${by}\:{roots}\:{of}\:{cubic}\:{equation} \\ $$$$\left({z}+\alpha{b}\right)^{\mathrm{3}} =\alpha^{\mathrm{3}_{} } \:\:\left(\alpha\neq\mathrm{0}\right). \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{Find}\:{product}\:{of}\:{all}\:{possible}\:{values} \\ $$$${of}\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}{i}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:.…
Question Number 49188 by behi83417@gmail.com last updated on 04/Dec/18 $${solve}\:{for}\:{x},{y},{z}\in{R}. \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{yz}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\boldsymbol{\mathrm{xz}}=\mathrm{2} \\ $$$$\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\boldsymbol{\mathrm{xy}}=\mathrm{3} \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 49186 by afachri last updated on 04/Dec/18 $$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{help}}… \\ $$$$ \\ $$$$\mathrm{There}\:\mathrm{is}\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{21}\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mathrm{21}\boldsymbol{{p}}\:\boldsymbol{{x}}\:+\:\mathrm{49}\boldsymbol{{p}}\:−\:\mathrm{7}\:=\:\mathrm{0} \\ $$$$\mathrm{whose}\:\mathrm{roots}\:\boldsymbol{{u}}\:\mathrm{and}\:\boldsymbol{{v}}.\:\mathrm{If}\:\boldsymbol{{u}}\:\mathrm{and}\:\boldsymbol{{v}}\:\mathrm{are}\:\mathrm{not}\:\in\mathbb{Z}\:,\: \\ $$$$\mathrm{and}\:\boldsymbol{{u}},\boldsymbol{{v}}\:\geqslant\:\mathrm{1}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{{u}}\:+\:\boldsymbol{{v}}\:! \\ $$$$…
Question Number 114720 by Algoritm last updated on 20/Sep/20 Answered by 1549442205PVT last updated on 20/Sep/20 $$\mathrm{x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{3}} +\mathrm{4x}−\mathrm{2}\Leftrightarrow\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{6x}−\mathrm{3}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 49177 by rahul 19 last updated on 04/Dec/18 $${If}\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \mid+\mid{z}_{\mathrm{2}} \mid\:,\:{then}\:{prove}\: \\ $$$${that}\:{arg}\left(\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\right)=\pi\:. \\ $$ Answered by MJS last…
Question Number 49174 by rahul 19 last updated on 04/Dec/18 $${If}\:{z}_{\mathrm{1}} ,{z}_{\mathrm{2}} \:{and}\:{z}_{\mathrm{3}} ,{z}_{\mathrm{4}\:} {are}\:{two}\:{pairs}\:{of}\: \\ $$$${conjugate}\:{complex}\:{numbers}\:,\:{then}\: \\ $$$${find}\:{value}\:{of}\:{arg}\left(\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{4}} }\right)+{arg}\left(\frac{{z}_{\mathrm{2}} }{{z}_{\mathrm{3}} }\right)\:? \\ $$…
Question Number 180236 by mnjuly1970 last updated on 09/Nov/22 $$\:\:\:\:\mathrm{calculation} \\ $$$$\:\:\:\:\Omega=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\zeta\:\left(\:\mathrm{2}{n}\:\right)}{\mathrm{4}^{\:{n}} }\:\overset{?} {=}\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:\Omega\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\mathrm{2}^{\:\mathrm{2}{n}} }\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\:\mathrm{2}{n}} }\:\right\} \\…
Question Number 49169 by rahul 19 last updated on 04/Dec/18 $${Find}\:{the}\:{least}\:{positive}\:{integer}\:{n}\:{such} \\ $$$${that}\:\left(\frac{\mathrm{2}{i}}{\mathrm{1}+{i}}\right)^{{n}} {is}\:{a}\:+{ve}\:{integer}\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18 $$\left(\frac{\mathrm{2}{i}\left(\mathrm{1}−{i}\right)}{\mathrm{2}}\right)^{{n}} \\…