Question Number 132635 by mohammad17 last updated on 15/Feb/21 $${where}\:{is}\:{to}\:{be}\:{the}\:{function}\: \\ $$$${f}\left({z}\right)={zRez}+\overset{−} {{z}Imz}+\overset{−} {{z}}\:{diffrentable}\:{and} \\ $$$${find}\:{the}\:{dervaitive}? \\ $$ Commented by mohammad17 last updated on 16/Feb/21…
Question Number 1559 by 123456 last updated on 19/Aug/15 $$\mathrm{find}\:\mathrm{complex}\:\mathrm{number}\:\alpha,\beta\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha^{{n}} =\beta^{{m}} \\ $$$$\beta^{{u}} =\alpha^{{v}} \\ $$$${n},{m},{u},{v}\in\mathbb{Z} \\ $$$$\boldsymbol{{Q}}\mathrm{1498} \\ $$ Answered by Rasheed…
Question Number 1557 by 123456 last updated on 19/Aug/15 $$\mathrm{find}\:\alpha,\beta,\gamma\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha=\beta^{\mathrm{2}} \\ $$$$\beta^{\mathrm{3}} =\gamma^{\mathrm{4}} \\ $$$$\gamma^{\mathrm{5}} =\alpha^{\mathrm{6}} \\ $$ Commented by 123456 last updated…
Question Number 132630 by Salman_Abir last updated on 15/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 132626 by mohammad17 last updated on 15/Feb/21 $${prove}\:{that}\:{Log}\left({z}_{\mathrm{1}} {z}_{\mathrm{2}} \right)={Logz}_{\mathrm{1}} +{Logz}_{\mathrm{2}} \\ $$$${with}\:{condition}\:−\pi<{Argz}_{\mathrm{1}} +{Argz}_{\mathrm{2}} <\pi \\ $$ Commented by mohammad17 last updated on…
Question Number 132620 by Tinku Tara last updated on 15/Feb/21 $$\mathrm{Reported}\:\mathrm{Issues} \\ $$$$\bullet\:\mathrm{text}\:\mathrm{extends}\:\mathrm{to}\:\mathrm{infinity} \\ $$$$\:\:\:\mathrm{problem}\:\mathrm{seen}.\:\mathrm{Will}\:\mathrm{fix}. \\ $$$$\bullet\:\mathrm{autosave}\:\mathrm{when}\:\mathrm{switching}\:\mathrm{app}. \\ $$$$\:\:\:\:\mathrm{this}\:\mathrm{is}\:\mathrm{already}\:\mathrm{present}\:\mathrm{so}\:\mathrm{if}\:\mathrm{you} \\ $$$$\:\:\:\:\mathrm{answer}\:\mathrm{a}\:\mathrm{phone}\:\mathrm{call}\:\mathrm{or}\:\mathrm{switch} \\ $$$$\:\:\:\:\mathrm{the}\:\mathrm{post}\:\mathrm{should}\:\mathrm{remain}.\: \\ $$$$\:\:\:\:\mathrm{will}\:\mathrm{check}\:\mathrm{again}\:\mathrm{and}\:\mathrm{address}.…
Question Number 132623 by Ashojaee79 last updated on 15/Feb/21 $${hi} \\ $$$${i}\:{am}\:{a}\:{sudent}.\:{i}\:{want}\:{remove} \\ $$$${watermak}\:{from}\:{output}\: \\ $$$${images}.\:{how}\:{can}\:{i}\:{use}\:{from} \\ $$$${premium}\:{options}? \\ $$ Commented by Tinku Tara last…
Question Number 67083 by lalitchand last updated on 22/Aug/19 $$\mathrm{CosA}+\mathrm{CosB}+\mathrm{CosC}=\mathrm{1}+\mathrm{4Cos}\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{2}}\right).\mathrm{Cos}\left(\frac{\mathrm{C}+\mathrm{A}}{\mathrm{2}}\right).\mathrm{Cos}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)=\mathrm{1}+\mathrm{4Cos}\left(\frac{\Pi−\mathrm{A}}{\mathrm{4}}\right).\mathrm{Cos}\left(\frac{\Pi−\mathrm{B}}{\mathrm{4}}\right).\mathrm{Cos}\left(\frac{\Pi−\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\Pi \\ $$ Answered by Tanmay chaudhury last updated on 23/Aug/19 $${LHS} \\ $$$$\mathrm{2}{cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right){cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 1545 by x-2 last updated on 17/Aug/15 $${x}−\mathrm{2}=\mathrm{0} \\ $$ Answered by 123456 last updated on 17/Aug/15 $${x}−\mathrm{2}=\mathrm{0} \\ $$$${x}−\mathrm{2}+\mathrm{2}=\mathrm{0}+\mathrm{2} \\ $$$${x}=\mathrm{2} \\…
Question Number 132613 by aurpeyz last updated on 15/Feb/21 $$\int{sin}^{−\mathrm{1}} {xdx} \\ $$ Answered by Dwaipayan Shikari last updated on 15/Feb/21 $${xsin}^{−\mathrm{1}} {x}−\int\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\…