Question Number 19733 by Tinkutara last updated on 15/Aug/17 $$\mathrm{If}\:\mid{z}\:+\:\mathrm{1}\mid\:=\:\sqrt{\mathrm{2}}\mid{z}\:−\:\mathrm{1}\mid,\:\mathrm{then}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{described}\:\mathrm{by}\:\mathrm{the}\:\mathrm{point}\:{z}\:\mathrm{in}\:\mathrm{the}\:\mathrm{argand} \\ $$$$\mathrm{diagram}\:\mathrm{is}\:\mathrm{a} \\ $$ Answered by ajfour last updated on 15/Aug/17 $$\mathrm{let}\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\…
Question Number 19732 by Tinkutara last updated on 15/Aug/17 $$\mathrm{The}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{given}\:\mathrm{by}\:\mid\frac{{z}\:−\:\mathrm{1}}{{z}\:−\:{i}}\mid\:=\:\mathrm{1}\:\mathrm{is} \\ $$ Answered by ajfour last updated on 15/Aug/17 $$\mathrm{z}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{perpendicular}\:\mathrm{bisector} \\ $$$$\mathrm{of}\:\mathrm{line}\:\mathrm{joining}\:\mathrm{z}=\mathrm{1}\:\mathrm{and}\:\mathrm{z}=\mathrm{i}\:\mathrm{that} \\ $$$$\:\:\mathrm{passes}\:\mathrm{through}\:\mathrm{origin}. \\…
Question Number 150801 by puissant last updated on 15/Aug/21 Commented by puissant last updated on 15/Aug/21 $${find}\:{tan}\theta\:{please}.. \\ $$ Commented by mr W last updated…
Question Number 19730 by Tinkutara last updated on 15/Aug/17 $$\mathrm{If}\:{z}\:=\:{x}\:+\:{iy}\:\mathrm{and}\:\mid{z}\:−\:\mathrm{2}{i}\mid\:=\:\mathrm{1},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{z}\:\mathrm{lies}\:\mathrm{on}\:{x}-\mathrm{axis} \\ $$$$\left(\mathrm{2}\right)\:{z}\:\mathrm{lies}\:\mathrm{on}\:{y}-\mathrm{axis} \\ $$$$\left(\mathrm{3}\right)\:{z}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$ Answered by ajfour last updated…
Question Number 19729 by Joel577 last updated on 15/Aug/17 $$\mathrm{2}{x}\:+\:\mathrm{9}{y}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:−\:\mathrm{45}{y}^{\mathrm{2}} \:+\:{xy}\:=\:\mathrm{0} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{xy} \\ $$ Answered by mrW1 last updated on…
Question Number 150797 by mathdanisur last updated on 15/Aug/21 $$\mathrm{x};\mathrm{y}\in\mathbb{N} \\ $$$$\frac{\mathrm{18}^{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{9}^{\boldsymbol{\mathrm{xy}}} }\:=\:\mathrm{2592}\:\:\Rightarrow\:\mathrm{find}\:\:\boldsymbol{\mathrm{xy}}=? \\ $$ Answered by nimnim last updated on 15/Aug/21…
Question Number 85262 by Rio Michael last updated on 20/Mar/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{symmetry}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\:\:\:{y}\:=\:\frac{\mathrm{1}}{{x}\:+\:\mathrm{2}} \\ $$ Commented by john santu last updated on 20/Mar/20 $$\mathrm{y}\:=−\:\mathrm{x}−\mathrm{2} \\…
Question Number 85260 by Umar last updated on 20/Mar/20 $$\mathrm{Evaluate}\:\mathrm{using}\:\mathrm{cauchy}'\mathrm{s}\:\mathrm{integral}\: \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{c}} \:\frac{\mathrm{e}^{\mathrm{i}\pi} }{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dz} \\ $$$$\mathrm{where}\:\mathrm{c}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mid\mathrm{z}−\mathrm{i}\mid=\mathrm{3}.\mathrm{5} \\ $$$$ \\ $$$$\mathrm{help}\:\mathrm{please} \\ $$…
Question Number 85261 by Umar last updated on 20/Mar/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}+\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$ \\ $$$$\mathrm{help}\:\mathrm{please} \\ $$ Commented by mathmax by abdo last updated on 20/Mar/20…
Question Number 150793 by cherokeesay last updated on 15/Aug/21 $$\left({R}−{r}\right)^{\mathrm{2}} \:+\:{R}^{\mathrm{2}} \:=\:\left({R}\:+\:{r}\right)^{\mathrm{2}} \:\Leftrightarrow \\ $$$${R}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} −\mathrm{2}{rR}\:+\:{R}^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} \:+\mathrm{2}{rR}\:\Rightarrow \\ $$$${R}^{\mathrm{2}} \:=\:\mathrm{4}{rR}\:\Rightarrow\:{r}\:=\:\frac{{R}}{\mathrm{4}}\:=\:\mathrm{1}{cm} \\ $$$$\mathscr{A}_{{S}}…