Question Number 20933 by Tinkutara last updated on 08/Sep/17 $$\mathrm{If}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{satisfying} \\ $$$${z}\:+\:{z}^{−\mathrm{1}} \:=\:\mathrm{1},\:\mathrm{then}\:{z}^{{n}} \:+\:{z}^{−{n}} ,\:{n}\:\in\:{N},\:\mathrm{has} \\ $$$$\mathrm{the}\:\mathrm{value} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}\left(−\mathrm{1}\right)^{{n}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of} \\ $$$$\mathrm{3}…
Question Number 20934 by Tinkutara last updated on 08/Sep/17 $$\mathrm{If}\:{z}\:\mathrm{satisfies}\:\mid{z}\:−\:\mathrm{1}\mid\:<\:\mid{z}\:+\:\mathrm{3}\mid,\:\mathrm{then}\:\omega\:= \\ $$$$\mathrm{2}{z}\:+\:\mathrm{3}\:−\:{i}\:\mathrm{satisfies} \\ $$$$\left(\mathrm{1}\right)\:\mid\omega\:−\:\mathrm{5}\:−\:{i}\mid\:<\:\mid\omega\:+\:\mathrm{3}\:+\:{i}\mid \\ $$$$\left(\mathrm{2}\right)\:\mid\omega\:−\:\mathrm{5}\mid\:<\:\mid\omega\:+\:\mathrm{3}\mid \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Im}\:\left({i}\omega\right)\:>\:\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\mid\mathrm{arg}\left(\omega\:−\:\mathrm{1}\right)\mid\:<\:\frac{\pi}{\mathrm{2}} \\ $$ Commented by Tinkutara…
Question Number 20932 by Tinkutara last updated on 08/Sep/17 $$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and}\:{z}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that},\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \\ $$$$=\:\mathrm{1}\:\mathrm{and}\:{b}\:+\:{ic}\:=\:\left(\mathrm{1}\:+\:{a}\right){z},\:\mathrm{then}\:\frac{\mathrm{1}\:+\:{iz}}{\mathrm{1}\:−\:{iz}} \\ $$$$\mathrm{equals}. \\ $$$$\left(\mathrm{1}\right)\:\frac{{b}\:−\:{ic}}{\mathrm{1}\:−\:{ia}} \\ $$$$\left(\mathrm{2}\right)\:\frac{{a}\:+\:{ib}}{\mathrm{1}\:+\:{c}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}\:−\:{c}}{{a}\:−\:{ib}} \\…
Question Number 151988 by mathdanisur last updated on 24/Aug/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}\:\:\mathrm{are}\:\mathrm{natural}\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\mathrm{2x}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{y}^{\boldsymbol{\mathrm{y}}} \:=\:\mathrm{3z}^{\boldsymbol{\mathrm{z}}} \\ $$$$\mathrm{then}\:\mathrm{find}\:\:\frac{\mathrm{2021}\boldsymbol{\mathrm{x}}\:+\:\mathrm{2022}\boldsymbol{\mathrm{y}}\:+\:\mathrm{2023}\boldsymbol{\mathrm{z}}}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}}\:=\:? \\ $$ Commented by mathdanisur last updated on 25/Aug/21…
Question Number 20914 by Tinkutara last updated on 07/Sep/17 $$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equation}\:{p}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{with} \\ $$$$\mathrm{real}\:\mathrm{coefficients}\:\mathrm{has}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{equation}\:{p}\left({p}\left({x}\right)\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Only}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{roots} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{All}\:\mathrm{real}\:\mathrm{roots} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Two}\:\mathrm{real}\:\mathrm{and}\:\mathrm{two}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots} \\…
Question Number 151979 by mathdanisur last updated on 24/Aug/21 $$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\mathrm{mx}\right)^{\boldsymbol{\mathrm{n}}} \:-\:\left(\mathrm{1}+\mathrm{nx}\right)^{\boldsymbol{\mathrm{m}}} }{\mathrm{x}^{\mathrm{2}} }\:=\:?\:\:;\:\:\mathrm{m};\mathrm{n}\in\mathbb{N} \\ $$ Answered by mr W last updated on 24/Aug/21 $$\left(\mathrm{1}+{mx}\right)^{{n}}…
Question Number 151956 by mathdanisur last updated on 24/Aug/21 $$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2k}^{\mathrm{2}} }\:\mathrm{arctan}\:\frac{\mathrm{2k}^{\mathrm{2}} \:-\:\mathrm{1}}{\mathrm{2k}^{\mathrm{2}} }\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 86426 by M±th+et£s last updated on 28/Mar/20 $${solve}\:{in}\:{R} \\ $$$${x}^{\mathrm{3}} −\mathrm{5}=\left[{x}\right] \\ $$ Answered by mr W last updated on 28/Mar/20 $${let}\:{x}={n}+\delta \\…
Question Number 20882 by soufiane zarik last updated on 06/Sep/17 Answered by $@ty@m last updated on 08/Sep/17 Commented by soufiane zarik last updated on 11/Sep/17…
Question Number 20881 by tawa tawa last updated on 05/Sep/17 Answered by mrW1 last updated on 06/Sep/17 $$\left(\mathrm{ax}+\mathrm{ay}\right)^{\mathrm{n}} =\mathrm{a}^{\mathrm{n}} \left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{n}} \\ $$$$\mathrm{since}\:\mathrm{a}\:\mathrm{middle}\:\mathrm{term}\:\mathrm{exists},\:\mathrm{n}\:\mathrm{is}\:\mathrm{even}. \\ $$$$\mathrm{its}\:\mathrm{middle}\:\mathrm{term}\:\mathrm{is} \\…