Question Number 131713 by mohammad17 last updated on 07/Feb/21 $${find}\:{all}\:{root}\:{by}\:{demover}\:{Z}^{\mathrm{4}} =\mathrm{2}−\mathrm{2}{i} \\ $$ Answered by mathmax by abdo last updated on 07/Feb/21 $$\mathrm{2}−\mathrm{2i}\:=\mathrm{2}\left(\mathrm{1}−\mathrm{i}\right)\:=\mathrm{2}\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:=\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}}…
Question Number 636 by 123456 last updated on 17/Feb/15 $${if}\:{f},{g}\:{are}\:{functions}\:{of}\:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${not}\:{constant}\:{such}\:{for}\:{all}\:\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$\begin{cases}{{f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)−{g}\left({x}\right){g}\left({y}\right)}\\{{g}\left({x}+{y}\right)={f}\left({x}\right){g}\left({y}\right)+{g}\left({x}\right){f}\left({y}\right)}\end{cases} \\ $$$${if}\:{f}'\left(\mathrm{0}\right)=\mathrm{0}\:{then}\:{proof}\:{os}\:{disproof} \\ $$$${that}\:\forall{x}\in\mathbb{R},\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} =\mathrm{1} \\ $$ Commented by prakash…
Question Number 131706 by bounhome last updated on 07/Feb/21 $$\int{e}^{\mathrm{3}{x}} {cosxdx}=\:?\:{help}\:{me}\:{please}\: \\ $$ Answered by mathmax by abdo last updated on 07/Feb/21 $$\int\:\mathrm{e}^{\mathrm{3x}} \:\mathrm{cosx}\:\mathrm{dx}\:=\mathrm{Re}\left(\int\:\mathrm{e}^{\mathrm{3x}} \:\mathrm{e}^{\mathrm{ix}}…
Question Number 131685 by aurpeyz last updated on 07/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 131684 by aurpeyz last updated on 07/Feb/21 Answered by ajfour last updated on 07/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\circleddash\left(−\mathrm{2}\mu{C}\:\right)\:\:\:\:\:\:\:\:\:\:\:\oplus\left(\mathrm{10}\mu{C}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{\mid{q}_{\mathrm{3}} {q}_{\mathrm{2}} \mid}{{r}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{\mid{q}_{\mathrm{3}} {q}_{\mathrm{1}}…
Question Number 131665 by naka3546 last updated on 07/Feb/21 Answered by mr W last updated on 07/Feb/21 Commented by mr W last updated on 07/Feb/21…
Question Number 596 by 123456 last updated on 08/Feb/15 $$\int\underset{{s}} {\int}\frac{{dx}\wedge{dy}+{dx}\wedge{dz}−{dy}\wedge{dz}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} } \\ $$$${where}\:{s}\:{is}\:{the}\:{surface}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1}\: \\ $$ Terms of Service Privacy…
Question Number 131654 by mohammad17 last updated on 07/Feb/21 $$\underset{\mathrm{0}} {\int}^{\:\mathrm{1}/\mathrm{64}} \frac{{tan}^{−\mathrm{1}} {x}}{\:\sqrt{{x}}}{dx} \\ $$ Answered by mathmax by abdo last updated on 07/Feb/21 $$\mathrm{I}=\int_{\mathrm{0}}…
Question Number 581 by 123456 last updated on 31/Jan/15 $${z}^{\mathrm{2}} =\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta \\ $$$${z}=? \\ $$ Answered by ssahoo last updated on 31/Jan/15 $${e}^{{i}\theta} =\mathrm{cos}\:\theta\:+{i}\mathrm{sin}\:\theta={z}^{\mathrm{2}} \\…
Question Number 66098 by aliesam last updated on 09/Aug/19 Commented by Prithwish sen last updated on 09/Aug/19 $$\left(\mathrm{cosx}+\boldsymbol{\mathrm{i}}\mathrm{sinx}\right)^{\mathrm{5}} =\mathrm{cos5x}+\boldsymbol{\mathrm{i}}\mathrm{sin5x}\:=\:\mathrm{cos}^{\mathrm{5}} \mathrm{x}+\mathrm{5}\boldsymbol{\mathrm{i}}\mathrm{cos}^{\mathrm{4}} \mathrm{xsinx}−\mathrm{10cos}^{\mathrm{3}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}−\mathrm{10}\boldsymbol{\mathrm{i}}\mathrm{cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{3}} \mathrm{x}+\mathrm{5cosxsin}^{\mathrm{4}}…