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Category: Relation and Functions

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Question Number 73411 by mathmax by abdo last updated on 11/Nov/19 $${calculate}\:\: \\ $$$$\left.\mathrm{1}\right){cos}\left(\mathrm{1}+{i}\right)\:,\:{sin}\left(\mathrm{1}+\mathrm{3}{i}\right) \\ $$$$\left.\mathrm{2}\right)\:{arctan}\left({i}\right),\:{arctan}\left(\mathrm{2}{i}\right)\:,\:{arctan}\left(\mathrm{1}+{i}\right)\:,{arctan}\left(\mathrm{1}−{i}\right)\:, \\ $$$${arctan}\left(\mathrm{1}+\mathrm{2}{i}\right). \\ $$$$\left.\mathrm{3}\right)\:{have}\:{us}\:\:{conj}\left({arctanz}\right)={arctan}\left(\overset{−} {{z}}\right)? \\ $$ Commented by…

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Question Number 7833 by mohitkumar88@gmail.com last updated on 18/Sep/16 $$\mathrm{3}\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{2}\frac{\mathrm{2}}{\mathrm{3}}= \\ $$ Answered by Rasheed Soomro last updated on 18/Sep/16 $$\mathrm{3}\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{2}\frac{\mathrm{2}}{\mathrm{3}}=\frac{\overset{\mathrm{5}} {\mathrm{15}}}{\underset{\mathrm{1}} {\mathrm{4}}}×\frac{\overset{\mathrm{2}} {\mathrm{8}}}{\underset{\mathrm{1}} {\mathrm{3}}}=\frac{\mathrm{10}}{\mathrm{1}}=\mathrm{10}…

let-w-x-0-lnt-x-2-t-2-2-dt-1-explicit-w-x-2-calculate-U-n-0-lnt-n-2-t-2-2-dt-find-lim-n-n-4-U-n-and-determine-nature-of-tbe-serie-U-n-

Question Number 73327 by mathmax by abdo last updated on 10/Nov/19 $${let}\:{w}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right)\:{explicit}\:{w}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left({n}^{\mathrm{2}} \:+{t}^{\mathrm{2}}…

Question-138829

Question Number 138829 by 676597498 last updated on 18/Apr/21 Answered by physicstutes last updated on 19/Apr/21 $${T}:\:{z}\rightarrow\omega\:\Rightarrow\:\omega\:=\:\mathrm{3}{z}\:+\:\mathrm{2}−\mathrm{5}{i} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{This}\:\mathrm{transformation}\:\mathrm{is}\:\mathrm{an}\:\mathrm{enlargment}\:\mathrm{of}\:\mathrm{scale}\:\mathrm{factor}\:\mathrm{3}\:\mathrm{followed}\:\mathrm{by}\: \\ $$$$\mathrm{a}\:\mathrm{translation}\:\mathrm{of}\:\left(\mathrm{2},−\mathrm{5}\right). \\ $$$$\left(\mathrm{b}\right)\:\mathrm{For}\:\mathrm{invariant}\:\mathrm{point},\:{f}\left({z}\right)={z} \\ $$$$\Rightarrow\:\:\:{z}\:=\:\mathrm{3}{z}+\:\mathrm{2}−\mathrm{5}{i}\:\:\Rightarrow\:{z}\:=\:−\mathrm{1}+\:\frac{\mathrm{5}}{\mathrm{2}}{i}…