Question Number 19629 by Tinkutara last updated on 13/Aug/17 $$\mathrm{If}\:\mid{z}\mid\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{the}\:\mathrm{points}\:\mathrm{representing} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:−\mathrm{1}\:+\:\mathrm{5}{z}\:\mathrm{will}\:\mathrm{lie} \\ $$$$\mathrm{on}\:\mathrm{a} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Straight}\:\mathrm{line} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Parabola} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Ellipse} \\ $$ Answered…
Question Number 19623 by Tinkutara last updated on 13/Aug/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{if}\:\mathrm{arg}\left(\frac{{z}\:−\:\mathrm{2}}{{z}\:−\:\mathrm{3}}\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$ Answered by ajfour last updated on 13/Aug/17 $$\mathrm{let}\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\Rightarrow\:\:\mathrm{arg}\left[\frac{\mathrm{x}−\mathrm{2}+\mathrm{iy}}{\mathrm{x}−\mathrm{3}+\mathrm{iy}}\right]=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{arg}\left[\frac{\left(\mathrm{x}−\mathrm{2}+\mathrm{iy}\right)\left(\mathrm{x}−\mathrm{3}−\mathrm{iy}\right)}{\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…
Question Number 150684 by mathdanisur last updated on 14/Aug/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in\mathbb{Z} \\ $$$$\left(\mathrm{b}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{d}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{b}\right)\left(\mathrm{d}-\mathrm{b}\right)\left(\mathrm{d}-\mathrm{c}\right) \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{is}\:\mathrm{divide} \\ $$$$\mathrm{into}\:\mathrm{12} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 19609 by ajfour last updated on 13/Aug/17 Answered by ajfour last updated on 13/Aug/17 $$\mathrm{Equation}\:\mathrm{of}\:\mathrm{L}_{\mathrm{1}} : \\ $$$$\:\:\:\:\:\:\bar {\mathrm{z}}_{\mathrm{1}} \mathrm{z}+\mathrm{z}_{\mathrm{1}} \bar {\mathrm{z}}−\mathrm{2}\mid\mathrm{z}_{\mathrm{1}} \mid^{\mathrm{2}}…
Question Number 19604 by ajfour last updated on 13/Aug/17 Commented by ajfour last updated on 13/Aug/17 $$\mathrm{solution}\:\mathrm{to}\:\mathrm{Q}.\mathrm{19508} \\ $$$$\mathrm{To}\:\mathrm{prove}\:\mathrm{p}=\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{0}} \mid=\mid\frac{\bar {\alpha}\mathrm{z}_{\mathrm{1}} +\alpha\bar {\mathrm{z}}_{\mathrm{1}} +\mathrm{2c}}{\mathrm{2}\mid\alpha\mid}\mid…
Question Number 19592 by ajfour last updated on 13/Aug/17 Commented by Tinkutara last updated on 13/Aug/17 $${z}_{{F}} \:=\:\frac{{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:{i}\frac{{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} } \\…
Question Number 150652 by mathdanisur last updated on 14/Aug/21 $$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{3x}+\mathrm{1}\right)+\mathrm{f}\left(\mathrm{5x}+\mathrm{1}\right)=\mathrm{x}^{\mathrm{3}} -\mathrm{2} \\ $$$$\mathrm{Find}\:\:\mathrm{f}\left(\mathrm{1}\right)+\mathrm{f}\left(\mathrm{4}\right)+\mathrm{f}\left(\mathrm{16}\right)=? \\ $$ Commented by MJS_new last updated on 14/Aug/21 $${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{152}}\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{153}\right)…
Question Number 150646 by mathdanisur last updated on 14/Aug/21 $$\boldsymbol{\mathrm{L}}\mathrm{et}\:\boldsymbol{\lambda}\in\mathbb{R}\:\mathrm{fixed}.\boldsymbol{\mathrm{S}}\mathrm{olve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\mathrm{ax}\:+\:\mathrm{by}\:=\:\mathrm{2}\lambda\:+\:\mathrm{1}}\\{\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{by}^{\mathrm{2}} \:=\:\mathrm{4}\lambda\:+\:\mathrm{1}}\\{\mathrm{ax}^{\mathrm{3}} \:+\:\mathrm{by}^{\mathrm{3}} \:=\:\mathrm{8}\lambda\:+\:\mathrm{1}}\\{\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{by}^{\mathrm{4}} \:=\:\mathrm{16}\lambda\:+\:\mathrm{1}}\end{cases} \\ $$ Answered by MJS_new last updated…
Question Number 19574 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Carol}\:\mathrm{was}\:\mathrm{given}\:\mathrm{three}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\mathrm{was}\:\mathrm{asked}\:\mathrm{to}\:\mathrm{add}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{three}\:\mathrm{to}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{two}. \\ $$$$\mathrm{Instead},\:\mathrm{she}\:\mathrm{multiplied}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{two},\:\mathrm{but}\:\mathrm{still}\:\mathrm{got} \\ $$$$\mathrm{the}\:\mathrm{right}\:\mathrm{answer}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{three}\:\mathrm{numbers}? \\ $$ Commented…
Question Number 85104 by jagoll last updated on 19/Mar/20 $$\mathrm{Given}\: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{3x}\:=\:−\mathrm{1}}\\{\mathrm{4y}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{6y}\:=\:−\mathrm{1}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{2y}\:−\:\mathrm{x} \\ $$ Commented by john santu last updated on…