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nice-calculus-suppose-z-x-iy-amp-z-1-3-p-iq-then-find-A-x-p-y-q-p-2-q-2-note-i-1-




Question Number 125096 by mnjuly1970 last updated on 08/Dec/20
            ... nice   calculus...      suppose :: z =x−iy  & (z)^(1/3)  =p+iq     then  find ::   A=(((x/p)+(y/q))/(p^2 +q^2 )) =??   note : i=(√(−1))
$$\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:{calculus}… \\ $$$$\:\:\:\:{suppose}\:::\:{z}\:={x}−{iy}\:\:\&\:\sqrt[{\mathrm{3}}]{{z}}\:={p}+{iq} \\ $$$$\:\:\:{then}\:\:{find}\:::\:\:\:{A}=\frac{\frac{{x}}{{p}}+\frac{{y}}{{q}}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\:=?? \\ $$$$\:{note}\::\:{i}=\sqrt{−\mathrm{1}} \\ $$
Answered by MJS_new last updated on 08/Dec/20
(z)^(1/3) =p+qi ⇒ z=p(p^2 −3q^2 )−q(q^2 −3p^2 ) ⇒  ⇒ x=p(p^2 −3q^2 )∧y=q(q^2 −3p^2 )  ⇒ A=−2
$$\sqrt[{\mathrm{3}}]{{z}}={p}+{q}\mathrm{i}\:\Rightarrow\:{z}={p}\left({p}^{\mathrm{2}} −\mathrm{3}{q}^{\mathrm{2}} \right)−{q}\left({q}^{\mathrm{2}} −\mathrm{3}{p}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\Rightarrow\:{x}={p}\left({p}^{\mathrm{2}} −\mathrm{3}{q}^{\mathrm{2}} \right)\wedge{y}={q}\left({q}^{\mathrm{2}} −\mathrm{3}{p}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{A}=−\mathrm{2} \\ $$
Commented by mnjuly1970 last updated on 08/Dec/20
 thanks alot sir MJS.grateful..
$$\:{thanks}\:{alot}\:{sir}\:{MJS}.{grateful}.. \\ $$

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