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If-a-vector-v-exists-in-n-dimensions-v-R-n-Can-there-exist-a-complex-dimension-s-

Question Number 6932 by FilupSmith last updated on 03/Aug/16 $$\mathrm{If}\:\mathrm{a}\:\mathrm{vector}\:\boldsymbol{{v}}\:\mathrm{exists}\:\mathrm{in}\:{n}\:\mathrm{dimensions}: \\ $$$$\boldsymbol{{v}}\in\mathbb{R}^{{n}} \\ $$$$\mathrm{Can}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{dimension}\left(\mathrm{s}\right)? \\ $$ Commented by nburiburu last updated on 04/Aug/16 $${you}\:{mean}\:{if}\:{v}\in\mathbb{R}^{{n}} \:\Rightarrow{v}\in\mathbb{C}^{{n}}…

If-z-x-is-a-complex-function-is-the-following-true-zdx-z-dx-i-z-dx-

Question Number 6737 by FilupSmith last updated on 19/Jul/16 $$\mathrm{If}\:{z}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{function}, \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{following}\:\mathrm{true}: \\ $$$$\int{zdx}=\int\Re\left({z}\right){dx}+{i}\int\Im\left({z}\right){dx} \\ $$ Commented by prakash jain last updated on 19/Jul/16 $$\mathrm{Yes}.\:\mathrm{Since}\:{i}\:\mathrm{is}\:\mathrm{simply}\:\mathrm{a}\:\mathrm{constant}.…

complex-analysis-if-f-n-x-d-n-dx-n-x-x-C-C-0-n-C-Z-0-and-g-n-x-0-1-f-n-x-d-then-find-the-value-of-

Question Number 137275 by mnjuly1970 last updated on 31/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:……{complex}\:\:{analysis}….. \\ $$$$\:\:\:\:{if}\:,\:\:{f}\left(\alpha,{n},{x}\right)=\frac{{d}^{\:{n}} }{{dx}^{{n}} }\left(\alpha^{{x}} \right)\:\:,\:{x}\in\mathbb{C} \\ $$$$\:\:\:\:\:\alpha\in\mathbb{C}−\left\{\mathrm{0}\right\}\:,\:{n}\in\mathbb{C}−\mathbb{Z}^{−} \cup\left\{\mathrm{0}\right\} \\ $$$$\:\:\:\:\:{and}\:\:{g}\left({n},{x}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left(\alpha,{n},{x}\right){d}\alpha \\ $$$$\:\:\:\:\:{then}\:\:{find}\:\:{the}\:{value}\:{of}\:… \\…

Given-that-one-of-the-value-for-the-5-th-root-of-a-complex-number-is-1-i-Find-the-another-four-values-

Question Number 136167 by ZiYangLee last updated on 19/Mar/21 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{value}\:\mathrm{for}\:\mathrm{the}\:\mathrm{5}^{\mathrm{th}} \\ $$$$\mathrm{root}\:\mathrm{of}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{is}\:−\mathrm{1}+{i}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{another}\:\mathrm{four}\:\mathrm{values}. \\ $$ Answered by mr W last updated on 19/Mar/21 $${one}\:\mathrm{5}^{{th}}…

calculate-0-e-z-2-dz-with-z-complex-

Question Number 136023 by mathmax by abdo last updated on 18/Mar/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } \mathrm{dz}\:\:\mathrm{with}\:\mathrm{z}\:\mathrm{complex} \\ $$ Commented by yutytfjh67ihd last updated on 25/Mar/21…

Hello-si-x-x-sin-x-x-dx-show-0-x-a-1-si-x-dx-a-sin-pia-2-a-hint-ipp-complex-Analysis-

Question Number 70361 by mind is power last updated on 03/Oct/19 $${Hello}\: \\ $$$${si}\left({x}\right)=−\int_{{x}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$$${show}\:\int_{\mathrm{0}} ^{+\infty} {x}^{{a}−\mathrm{1}} {si}\left({x}\right){dx}=−\frac{\Gamma\left({a}\right){sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}{{a}} \\ $$$${hint}\:{ipp}\:+{complex}\:{Analysis} \\ $$…

soit-E-l-equation-complex-z-2-2pz-q-0-p-q-C-let-u-C-u-2-q-show-that-if-z-1-z-2-are-solutions-of-E-z-1-z-2-p-u-p-u-

Question Number 134461 by pticantor last updated on 04/Mar/21 $${soit}\:\left(\boldsymbol{{E}}\right)\:{l}'{equation}\:{complex}: \\ $$$$\boldsymbol{{z}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{pz}}−\boldsymbol{{q}}=\mathrm{0} \\ $$$$\boldsymbol{{p}},\boldsymbol{{q}}\in\mathbb{C} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}\in\mathbb{C}\backslash\:\boldsymbol{{u}}^{\mathrm{2}} =\boldsymbol{{q}} \\ $$$$\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{if}}\:\boldsymbol{{z}}_{\mathrm{1}} ,\boldsymbol{{z}}_{\mathrm{2}} \boldsymbol{{are}}\:\boldsymbol{{solutions}}\:\boldsymbol{{of}}\:\left(\boldsymbol{{E}}\right), \\ $$$$\mid\boldsymbol{{z}}_{\mathrm{1}} \mid+\mid\boldsymbol{{z}}_{\mathrm{2}}…

why-the-function-cosz-and-sinz-it-is-not-bounded-in-complex-number-

Question Number 134419 by mohammad17 last updated on 03/Mar/21 $${why}\:{the}\:{function}\:{cosz}\:{and}\:{sinz}\:{it}\:{is}\: \\ $$$${not}\:{bounded}\:{in}\:{complex}\:{number}\:? \\ $$ Answered by Olaf last updated on 03/Mar/21 $$\mathrm{cos}{z}\:=\:\mathrm{cos}\left({x}+{iy}\right) \\ $$$$\mathrm{cos}{z}\:=\:\mathrm{cos}{x}\mathrm{cos}\left({iy}\right)−\mathrm{sin}{x}\mathrm{sin}\left({iy}\right) \\…