Question Number 84564 by Rio Michael last updated on 14/Mar/20 $$\mathrm{find}\:\mathrm{arg}\left(\mathrm{z}\right)\:\mathrm{given}\:\mathrm{that}\:\:\mathrm{z}\:=\:\frac{\mathrm{1}\:+\:{i}}{\mathrm{1}−{i}} \\ $$ Commented by mr W last updated on 14/Mar/20 $${z}=\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}=\frac{\left(\mathrm{1}+{i}\right)\left(\mathrm{1}+{i}\right)}{\mathrm{1}−{i}^{\mathrm{2}} }=\frac{\mathrm{1}+\mathrm{2}{i}+{i}^{\mathrm{2}} }{\mathrm{2}}={i} \\…
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Question Number 84558 by Kunal12588 last updated on 14/Mar/20 $${happy}\:\pi\:{day} \\ $$ Commented by john santu last updated on 14/Mar/20 $$\pi\:−\:\mathrm{day}\:\mathrm{happy} \\ $$ Terms of…
Question Number 84553 by naka3546 last updated on 14/Mar/20 $${Find}\:\:{mininum}\:\:{value}\:\:{of}\:\:{n}\:\:{such}\:\:{that} \\ $$$${both}\:\:{n}\:+\:\mathrm{3}\:\:\:{and}\:\:\mathrm{2020}{n}\:+\:\mathrm{1}\:\:{are}\:\:{square}\:\:{numbers}\:. \\ $$ Commented by mr W last updated on 14/Mar/20 $${i}\:{got}\:{n}_{{min}} =\mathrm{2022} \\…
Question Number 150062 by 0731619 last updated on 09/Aug/21 Answered by Ar Brandon last updated on 09/Aug/21 $${li}_{\mathrm{2}} \left({x}\right)\underset{{function}} {\overset{{dilogarithm}} {=}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }=−\int_{\mathrm{0}}…
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Question Number 84496 by bshahid010@gmail.com last updated on 13/Mar/20 Commented by mathmax by abdo last updated on 13/Mar/20 $${I}\:=\int\:\:\frac{{arcsinx}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts}\:\:{u}^{'} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{and}\:{v}={arcsinx}\:\Rightarrow \\ $$$${I}\:=−\frac{{arsinx}}{{x}}\:−\int−\frac{\mathrm{1}}{{x}}\:×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}…
Question Number 149996 by tabata last updated on 08/Aug/21 $$\left(\mathrm{1}\right)\:\int\:\:\frac{{dx}}{\mathrm{1}+{tanx}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\int\:\:\frac{\sqrt{{tanx}}}{{sinx}\:{cosx}}{dx} \\ $$ Answered by mindispower last updated on 08/Aug/21 $$=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}.\frac{\sqrt{{tg}\left({x}\right)}}{{tg}\left({x}\right)}{dx}=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}}…
Question Number 149989 by ZiYangLee last updated on 08/Aug/21 $$\mathrm{By}\:\mathrm{subs}\:{u}^{\mathrm{2}} =\mathrm{4}+{x},\:\mathrm{evaluate}\:\int\:\frac{\sqrt{\mathrm{4}+{x}}}{{x}}\:{dx} \\ $$ Answered by puissant last updated on 08/Aug/21 $$\int\frac{\sqrt{\mathrm{4}+{x}}}{{x}}{dx}={Q} \\ $$$${u}=\sqrt{\mathrm{4}+{x}}\:\rightarrow\:{u}^{\mathrm{2}} =\mathrm{4}+{x}\:\rightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{4}…
Question Number 149959 by Mathfinity last updated on 08/Aug/21 Commented by john_santu last updated on 08/Aug/21 Commented by john_santu last updated on 08/Aug/21 Commented by…