Question Number 130818 by Study last updated on 29/Jan/21 $${if}\:\mid{z}_{\mathrm{1}} \mid=\mid{z}_{\mathrm{2}} \mid=\mid{z}_{\mathrm{3}} \mid=\mathrm{1}\:\: \\ $$$${and}\:\:\:\mid\frac{\mathrm{1}}{{z}_{\mathrm{1}} }\mid+\mid\frac{\mathrm{1}}{{z}_{\mathrm{2}} }\mid+\mid\frac{\mathrm{1}}{{z}_{\mathrm{3}} }\mid=\mathrm{1} \\ $$$$\:{find}\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \mid=?\:\:\: \\ $$$${z}_{\mathrm{1}}…
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Question Number 130753 by Algoritm last updated on 28/Jan/21 Answered by Dwaipayan Shikari last updated on 28/Jan/21 $$\left.{a}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }=\mathrm{4}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{2}^{{n}} }−\overset{\infty} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}}…
Question Number 65214 by naka3546 last updated on 26/Jul/19 $${Prove}\:\:{or}\:\:{disprove}\:\:\:{that}\:\:\:\mathrm{2}^{\mathrm{101}} \:\mid\:{n}^{{n}} \:−\:\mathrm{101}\:. \\ $$ Commented by naka3546 last updated on 26/Jul/19 $${n}\:\:{positive}\:\:{integer} \\ $$ Commented…
Question Number 130744 by rs4089 last updated on 28/Jan/21 Answered by Dwaipayan Shikari last updated on 28/Jan/21 $${n}+\mathrm{1}=\sqrt{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}}=\sqrt{\mathrm{1}+{n}\left({n}+\mathrm{2}\right)} \\ $$$$=\sqrt{\mathrm{1}+{n}\sqrt{\mathrm{1}+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)}}=\sqrt{\mathrm{1}+{n}\sqrt{\mathrm{1}+\left({n}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left({n}+\mathrm{2}\right)\sqrt{\mathrm{1}+…}}}} \\ $$$${n}=\mathrm{2} \\ $$$$\mathrm{3}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}+…}}}}…
Question Number 130745 by greg_ed last updated on 28/Jan/21 $$\mathrm{hi},\:\mathrm{mr}\:\mathrm{tinku}\:\mathrm{tara}\:! \\ $$$$\mathrm{who}\:\mathrm{must}\:\mathrm{report}\:\mathrm{an}\:\mathrm{inappropriate}\:\mathrm{post} \\ $$$$\mathrm{on}\:\mathrm{this}\:\mathrm{forum}\:??? \\ $$$$\mathrm{should}\:\mathrm{any}\:\mathrm{member}\:\mathrm{of}\:\mathrm{the}\:\mathrm{forum}\:\mathrm{have}\: \\ $$$$\mathrm{this}\:\mathrm{right}\:? \\ $$$$\mathrm{does}\:\mathrm{this}\:\mathrm{forum}\:\mathrm{have}\:\mathrm{a}\:\mathrm{moderator}\:? \\ $$ Commented by Ar…
Question Number 130742 by mohammad17 last updated on 28/Jan/21 Answered by Dwaipayan Shikari last updated on 28/Jan/21 $${z}=\mathrm{9}+\mathrm{3}{i}\:\:=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }\:{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{3}\sqrt{\mathrm{10}}{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${z}=\mathrm{7}+\mathrm{2}{i}=\sqrt{\mathrm{53}}\:{e}^{{itan}^{−\mathrm{1}}…
Question Number 65192 by naka3546 last updated on 26/Jul/19 $${Minimum}\:\:{value}\:\:{of} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mid\:\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}\:+\:\mathrm{tan}\:{x}\:+\:\mathrm{cot}\:{x}\:+\:\mathrm{sec}\:{x}\:+\:\mathrm{cosec}\:{x}\:\mid\:\:\:{is}\:\:… \\ $$ Answered by Tanmay chaudhury last updated on 26/Jul/19 $${sinx}+{cosx}\:\rightarrow\left({let}\:{sinx}+{cosx}={a}\right) \\ $$$$\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sinx}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{cosx}\right)…