Question Number 182423 by Noorzai last updated on 09/Dec/22 Answered by JDamian last updated on 09/Dec/22 $${KK}=\mathrm{11}{k} \\ $$$${LL}=\mathrm{11}{l} \\ $$$$ \\ $$$${KK}×{LL}=\mathrm{11}^{\mathrm{2}} {kl}=\mathrm{4235} \\…
Question Number 51321 by Tawa1 last updated on 25/Dec/18 $$\mathrm{A}\:\mathrm{B}\:\mathrm{C}\:\mathrm{D}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{parallelogram}\:\mathrm{on}\:\mathrm{the}\:\mathrm{Argand}\:\mathrm{plane}.\:\mathrm{The}\: \\ $$$$\mathrm{affixes}\:\mathrm{of}\:\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\:\mathrm{are}\:\:\:\mathrm{8}\:+\:\mathrm{5i},\:\:−\:\mathrm{7}\:−\:\mathrm{5i},\:\:−\:\mathrm{5}\:+\:\mathrm{5i}\:\:\mathrm{respectively} \\ $$$$.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{affix}\:\mathrm{of}\:\:\mathrm{D} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18 $${A}\left(\mathrm{8},\mathrm{5}\right)\:\:{B}\left(−\mathrm{7},−\mathrm{5}\right)\:{C}\left(−\mathrm{5},\mathrm{5}\right)\:{D}\left({x},{y}\right) \\…
Question Number 51320 by Tawa1 last updated on 25/Dec/18 $$\mathrm{The}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\:\mathrm{represent}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:\:\mathrm{z}_{\mathrm{1}} ,\:\mathrm{z}_{\mathrm{2}} ,\:\mathrm{z}_{\mathrm{3}} \: \\ $$$$\mathrm{respectively}.\:\mathrm{And}\:\mathrm{G}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centroid}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{A}\:\mathrm{B}\:\mathrm{C},\:\:\mathrm{if} \\ $$$$\mathrm{4z}_{\mathrm{1}} \:+\:\mathrm{z}_{\mathrm{2}} \:+\:\mathrm{z}_{\mathrm{3}} \:\:=\:\:\mathrm{0},\:\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mid}\:\mathrm{point}\:\mathrm{of}\:\:\mathrm{AG}. \\ $$ Answered by tanmay.chaudhury50@gmail.com…
Question Number 182394 by Tawa11 last updated on 08/Dec/22 Commented by Tawa11 last updated on 08/Dec/22 $$\mathrm{Number}\:\mathrm{18} \\ $$ Answered by CrispyXYZ last updated on…
Question Number 51316 by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 Commented by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 Commented by Cheyboy last updated on 26/Dec/18 $$\mathrm{Waaw}!!\:\mathrm{this}\:\mathrm{is}\:\mathrm{very}\:\:\mathrm{good}.\mathrm{thank} \\…
Question Number 51312 by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 Commented by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 51314 by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 Commented by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 51307 by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 Commented by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 $${this}\:{is}\:{not}\:{question}\:{but}\:{may}\:{help}\:{others}… \\ $$ Commented by afachri last updated on…
Question Number 116822 by bemath last updated on 07/Oct/20 $$\:\mathrm{what}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\sqrt{{i}}\:=? \\ $$ Commented by Dwaipayan Shikari last updated on 07/Oct/20 $$\sqrt{{i}}=\left(\sqrt{\frac{\mathrm{2}{i}}{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{\left(\mathrm{1}+\mathrm{2}{i}+{i}^{\mathrm{2}} \right)}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$ Answered…
Question Number 51284 by Tawa1 last updated on 25/Dec/18 $$\mathrm{If}\:\:\boldsymbol{\mathrm{x}}\:\mathrm{is}\:\mathrm{real},\:\mathrm{show}\:\mathrm{that}\:\:\left(\mathrm{2}\:+\:\mathrm{j}\right)\mathrm{e}^{\left(\mathrm{1}\:+\:\mathrm{j3}\right)\boldsymbol{\mathrm{x}}} \:+\:\left(\mathrm{2}\:−\:\boldsymbol{\mathrm{j}}\right)\boldsymbol{\mathrm{e}}^{\left(\mathrm{1}\:−\:\boldsymbol{\mathrm{j}}\mathrm{3}\right)\boldsymbol{\mathrm{x}}} \\ $$$$\mathrm{is}\:\mathrm{also}\:\mathrm{real} \\ $$ Commented by maxmathsup by imad last updated on 25/Dec/18 $${first}\:{what}\:{mean}\:{j}\:{and}\:{j}\mathrm{3}?…