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Question-130188

Question Number 130188 by SEKRET last updated on 23/Jan/21 Answered by Lordose last updated on 23/Jan/21 $$ \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}} }\mathrm{dx}\:=\:\boldsymbol{\beta}\left(\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\right)\:=\:\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\mathrm{2}\right)}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$…

if-Z-is-any-boint-on-the-circle-Z-1-1-prove-that-arg-Z-1-2arg-Z-2-3-arg-Z-2-Z-

Question Number 130186 by mohammad17 last updated on 23/Jan/21 $${if}\:{Z}\:{is}\:{any}\:{boint}\:{on}\:{the}\:{circle}\:\mid{Z}−\mathrm{1}\mid=\mathrm{1} \\ $$$$ \\ $$$${prove}\:{that}\:{arg}\left({Z}−\mathrm{1}\right)=\mathrm{2}{arg}\left({Z}\right)=\frac{\mathrm{2}}{\mathrm{3}}{arg}\left({Z}^{\mathrm{2}} −{Z}\right) \\ $$ Commented by mohammad17 last updated on 23/Jan/21 $$?????…

log2-X-log-2-2X-log-2-4X-X-

Question Number 130168 by Adel last updated on 23/Jan/21 $$\mathrm{log}\underset{\mathrm{X}} {\mathrm{2}}\centerdot\mathrm{log}\:\underset{\mathrm{2X}} {\mathrm{2}}=\mathrm{log}\:\underset{\mathrm{4X}} {\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{X}=? \\ $$ Commented by liberty last updated on 23/Jan/21 $$\mathrm{do}\:\mathrm{you}\:\mathrm{meant}\:\mathrm{log}\:_{{x}} \left(\mathrm{2}\right).\mathrm{log}\:_{\mathrm{2}{x}} \left(\mathrm{2}\right)\:=\:\mathrm{log}\:_{\mathrm{4}{x}}…

if-U-x-y-2x-x-3-3xy-2-construct-a-complex-analytical-function-the-expression-of-which-is-the-true-part-

Question Number 130102 by mohammad17 last updated on 22/Jan/21 $${if}\:{U}\left({x},{y}\right)=\mathrm{2}{x}−{x}^{\mathrm{3}} +\mathrm{3}{xy}^{\mathrm{2}} \:{construct}\:{a}\:{complex}\: \\ $$$${analytical}\:{function},\:{the}\:{expression} \\ $$$${of}\:{which}\:{is}\:{the}\:{true}\:{part}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

find-the-key-value-Arg-z-for-2-i-1-i-

Question Number 130104 by mohammad17 last updated on 22/Jan/21 $${find}\:{the}\:{key}\:{value}\:\left[{Arg}\left({z}\right)\right]{for}\:\left(\mathrm{2}+{i}\right)^{\mathrm{1}−{i}} \\ $$ Answered by Olaf last updated on 22/Jan/21 $${z}\:=\:\left(\mathrm{2}+{i}\right)^{\mathrm{1}−{i}} \\ $$$$\mathrm{Z}\:=\:\mathrm{ln}{z}\:=\:\left(\mathrm{1}−{i}\right)\mathrm{ln}\left(\mathrm{2}+{i}\right) \\ $$$$\mathrm{Z}\:=\:\left(\mathrm{1}−{i}\right)\mathrm{ln}\left(\sqrt{\mathrm{5}}{e}^{{i}\mathrm{arctan}\frac{\mathrm{1}}{\mathrm{2}}} \right)…