Question Number 25139 by tawa tawa last updated on 05/Dec/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{real}\:\mathrm{and}\:\mathrm{the}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{number}\:\:\:\mathrm{z}\:=\:\left(−\:\mathrm{1}\right)^{\mathrm{1000003}} \\ $$ Answered by Rasheed.Sindhi last updated on 05/Dec/17 $$\left(−\mathrm{i}\right)^{\mathrm{1000003}} =\left\{\left(−\mathrm{i}\right)^{\mathrm{2}} \right\}^{\mathrm{500001}} \left(−\mathrm{i}\right) \\…
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Question Number 25066 by Mr easy last updated on 02/Dec/17 $${let}\:{a},{b},{c},{x},{y}\:{and}\:{z}\:{be}\:{complex}\:{number} \\ $$$${such}\:{that}\:{a}=\frac{{b}+{c}}{{x}−\mathrm{2}}\:,{b}=\frac{{c}+{a}}{{y}−\mathrm{2}}\:\:\:\:{c}=\frac{{a}+{b}}{{z}−\mathrm{2}}. \\ $$$${xy}\:+{yz}\:+{zx}=\mathrm{1000}\:{and}\:{x}+{y}+{z}=\mathrm{2016} \\ $$$${find}\:{the}\:{value}\:{of}\:{xyz}. \\ $$ Answered by ajfour last updated on…
Question Number 89937 by ~blr237~ last updated on 20/Apr/20 $${Prove}\:{that}\:{for}\:{all}\:{complex}\:{such}\:{as}\:\mid{z}\mid<\mathrm{1}= \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{{n}} }{\left({z}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{nz}^{{n}} }{{z}^{{n}} −\mathrm{1}}\:=\:\mathrm{0}\: \\ $$ Commented by…
Question Number 23146 by Tinkutara last updated on 26/Oct/17 $$\mathrm{Square}\:\mathrm{planar}\:\mathrm{complex}\:\mathrm{is}\:\mathrm{formed}\:\mathrm{by} \\ $$$$\mathrm{hybridisation}\:\mathrm{of}\:\mathrm{which}\:\mathrm{atomic}\:\mathrm{orbitals}? \\ $$$$\left(\mathrm{1}\right)\:{s},\:{p}_{{x}} ,\:{p}_{{y}} ,\:{p}_{{z}} \\ $$$$\left(\mathrm{2}\right)\:{s},\:{p}_{{x}} ,\:{p}_{{y}} ,\:{d}_{{z}^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:{s},\:{p}_{{x}} ,\:{p}_{{y}} ,\:{d}_{{x}^{\mathrm{2}}…
Question Number 88314 by M±th+et£s last updated on 09/Apr/20 $$\left(\:{a},{b}\:\right){are}\:{complex}\:{numbers}\:{and}\:{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:\left(\frac{{a}}{{a}+{b}}\right)^{\mathrm{2020}} +\left(\frac{{b}}{{a}+{b}}\right)^{\mathrm{2020}} \\ $$$$ \\ $$ Commented by mathmax by abdo last…
Question Number 22387 by gopikrishnan005@gmail.com last updated on 17/Oct/17 $${the}\:{arguments}\:{of}\:{n}^{{th}} \:{roots}\:{of}\:{a}\:{complex}\:{number}\:{differ}\:{by} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 87497 by Rio Michael last updated on 04/Apr/20 $$\mathrm{A}\:\mathrm{complex}\:\mathrm{number}\:{z}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:{z}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\theta\:+\:{i}\mathrm{sin}\:\theta\right),\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{z}^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\left(\mathrm{cos}\:{n}\theta\:+\:{i}\mathrm{sin}\:{n}\theta\right) \\ $$$$\mathrm{Using}\:\mathrm{De}\:\mathrm{Moivre}'\mathrm{s}\:\mathrm{theorem},\mathrm{or}\:\mathrm{otherwise},\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\left(\mathrm{i}\right)\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{4}^{{r}} }\:\mathrm{sin}\:\mathrm{2r}\theta\:\mathrm{is}\:\mathrm{a}\:\mathrm{convergent}\:\mathrm{geometic}\:\mathrm{progression}. \\ $$$$\left(\mathrm{ii}\right)\:\underset{{r}=\mathrm{0}} {\overset{\infty}…
Question Number 87306 by abdomathmax last updated on 03/Apr/20 $${calculate}\:{by}\:{complex}\:{method}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{xdx}}{{x}^{\mathrm{4}} \:+\mathrm{1}} \\ $$ Commented by Ar Brandon last updated on 03/Apr/20 $${My}\:\:{suggestion} \\…
Question Number 152757 by EDWIN88 last updated on 01/Sep/21 $${Find}\:{all}\:{complex}\:{number}\:{z}\:{such} \\ $$$${that}\:\left(\mathrm{3}{z}+\mathrm{1}\right)\left(\mathrm{4}{z}+\mathrm{1}\right)\left(\mathrm{6}{z}+\mathrm{1}\right)\left(\mathrm{12}{z}+\mathrm{1}\right)=\mathrm{2} \\ $$ Answered by john_santu last updated on 01/Sep/21 $${note}\:{that}\:\mathrm{8}\left(\mathrm{3}{z}+\mathrm{1}\right)\mathrm{6}\left(\mathrm{4}{z}+\mathrm{1}\right)\mathrm{4}\left(\mathrm{6}{z}+\mathrm{1}\right)\mathrm{2}\left(\mathrm{12}{z}+\mathrm{1}\right)=\mathrm{768} \\ $$$$\left(\mathrm{24}{z}+\mathrm{8}\right)\left(\mathrm{24}{z}+\mathrm{6}\right)\left(\mathrm{24}{z}+\mathrm{4}\right)\left(\mathrm{24}{z}+\mathrm{2}\right)=\mathrm{768} \\…
Question Number 21308 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{complex}\:\mathrm{numbers}\:{z}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mid{z}\:−\:\mid{z}\:+\:\mathrm{1}\mid\mid\:=\:\mid{z}\:+\:\mid{z}\:−\:\mathrm{1}\mid\mid \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com