Question Number 41519 by maxmathsup by imad last updated on 08/Aug/18 $${let}\:{z}=\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\:−{i}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${calculate}\:\mid{z}^{{n}} \mid\:\:{and}\:{arg}\left({z}^{{n}} \right) \\ $$ Answered by alex041103 last updated on 09/Aug/18…
Question Number 41520 by maxmathsup by imad last updated on 08/Aug/18 $${let}\:{Z}\:=\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:+{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:{and}\:\:{A}=\:{Z}+{Z}^{\mathrm{2}} \:+{Z}^{\mathrm{4}} \\ $$$${B}={Z}^{\mathrm{3}} \:+{Z}^{\mathrm{5}} \:+{Z}^{\mathrm{6}} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\overset{−} {{A}}={B} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{A}+{B}\:=−\mathrm{1}\:{and}\:{A}.{B}\:=\mathrm{2} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\:{A}\:{and}\:{B}. \\…
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Question Number 172560 by mr W last updated on 28/Jun/22 $${prove}\:{for}\:{n}\in{N}\:{and}\:{n}>\mathrm{1} \\ $$$$\left(\frac{{n}+\mathrm{1}}{\mathrm{3}}\right)^{{n}} <{n}!<\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$ Commented by mr W last updated on 07/Sep/22 $${see}\:{also}\:{Q}\mathrm{173815}…
Question Number 172532 by Mikenice last updated on 28/Jun/22 Answered by MJS_new last updated on 28/Jun/22 $${x}_{\mathrm{1}} \approx−\mathrm{4}.\mathrm{24290} \\ $$$${x}_{\mathrm{2}} \approx.\mathrm{0593074} \\ $$$${x}_{\mathrm{3}} =\mathrm{3} \\…
Question Number 172516 by Mikenice last updated on 28/Jun/22 $${solve} \\ $$$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} \\ $$ Commented by mr W last updated on 28/Jun/22 $${Q}#\:\mathrm{172029} \\…
Question Number 172513 by Mikenice last updated on 28/Jun/22 $${simplify} \\ $$$$\sqrt{{a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} +\mathrm{2}{ab}^{\mathrm{2}} }\:−\sqrt{{a}^{\mathrm{2}} {b}+\mathrm{4}{b}^{\mathrm{3}} +\mathrm{4}{ab}^{\mathrm{3}} } \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 172515 by Mikenice last updated on 28/Jun/22 $${simplify} \\ $$$$\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}−\mathrm{1}}\boldsymbol{\div}\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}×\frac{\mathrm{1}}{{z}+\frac{\mathrm{1}}{{z}}} \\ $$ Answered by Rasheed.Sindhi last updated on 28/Jun/22 $$\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}−\mathrm{1}}\boldsymbol{\div}\frac{\mathrm{1}}{{z}^{\mathrm{2}}…