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Question Number 49200 by rahul 19 last updated on 04/Dec/18
1)Find the area of the triangle formed by roots of cubic equation (z+αb)^3 =α^3_ (α≠0). 2) Find product of all possible values of ((1/2)+(((√3)i)/2))^(3/4) .
$$\left.\mathrm{1}\right){Find}\:{the}\:{area}\:{of}\:{the}\:{triangle}\:{formed} \\ $$$${by}\:{roots}\:{of}\:{cubic}\:{equation} \\ $$$$\left({z}+\alpha{b}\right)^{\mathrm{3}} =\alpha^{\mathrm{3}_{} } \:\:\left(\alpha\neq\mathrm{0}\right). \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{Find}\:{product}\:{of}\:{all}\:{possible}\:{values} \\ $$$${of}\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}{i}}{\mathrm{2}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:. \\ $$
Commented by rahul 19 last updated on 04/Dec/18
Ans→1) 1.3 α^2 ... 2) 1 ...
$$\left.{Ans}\rightarrow\mathrm{1}\right)\:\mathrm{1}.\mathrm{3}\:\alpha^{\mathrm{2}} … \\ $$$$\left.\mathrm{2}\right)\:\mathrm{1}\:… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
2)another aporoach.. ((1/2)+((√3)/2)i)^(3/4) =(e^(i(π/3)) )^(3/4) =(e^(iπ) )^(1/4) =(cos2kπ+isin2kπ)^(1/4) =(cos((2kπ)/4)+isin((2kπ)/4))=e^((i2kπ)/4) so required ans is e^(i×((2×0×π)/4)) ×e^(i((2×1×π)/4)) ×e^(i((2×−1×π)/4)) ×e^(i((2×2×π)/4)) =e^0 ×e^0 ×e^(iπ) =1×1×(cosπ+isinπ)=1
$$\left.\mathrm{2}\right){another}\:{aporoach}.. \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} =\left({e}^{{i}\frac{\pi}{\mathrm{3}}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} =\left({e}^{{i}\pi} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} =\left({cos}\mathrm{2}{k}\pi+{isin}\mathrm{2}{k}\pi\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\left({cos}\frac{\mathrm{2}{k}\pi}{\mathrm{4}}+{isin}\frac{\mathrm{2}{k}\pi}{\mathrm{4}}\right)={e}^{\frac{{i}\mathrm{2}{k}\pi}{\mathrm{4}}} \\ $$$${so}\:{required}\:{ans}\:{is} \\ $$$${e}^{{i}×\frac{\mathrm{2}×\mathrm{0}×\pi}{\mathrm{4}}} ×{e}^{{i}\frac{\mathrm{2}×\mathrm{1}×\pi}{\mathrm{4}}} ×{e}^{{i}\frac{\mathrm{2}×−\mathrm{1}×\pi}{\mathrm{4}}} ×{e}^{{i}\frac{\mathrm{2}×\mathrm{2}×\pi}{\mathrm{4}}} \\ $$$$={e}^{\mathrm{0}} ×{e}^{\mathrm{0}} ×{e}^{{i}\pi} =\mathrm{1}×\mathrm{1}×\left({cos}\pi+{isin}\pi\right)=\mathrm{1} \\ $$
Commented by rahul 19 last updated on 05/Dec/18
cos π =−1 , right? So your answer comes out to be −1... Thank you sir for providing hint i ′ve got the correct ans. now :)
$$\mathrm{cos}\:\pi\:=−\mathrm{1}\:,\:{right}? \\ $$$${So}\:{your}\:{answer}\:{comes}\:{out}\:{to}\:{be}\:−\mathrm{1}… \\ $$$${Thank}\:{you}\:{sir}\:{for}\:{providing}\:{hint}\:{i}\:'{ve}\: \\ $$$$\left.{got}\:{the}\:{correct}\:{ans}.\:{now}\::\right) \\ $$
Commented by rahul 19 last updated on 05/Dec/18
You get z= ( e^(iπ) )^(1/4) =(−1)^(1/4) ⇒z^4 +1=0 ∴ Product = 1.
$${You}\:{get}\: \\ $$$${z}=\:\left(\:{e}^{{i}\pi} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} =\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\Rightarrow{z}^{\mathrm{4}} +\mathrm{1}=\mathrm{0}\: \\ $$$$\therefore\:{Product}\:=\:\mathrm{1}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
yes cosπ=−1 typo..
$${yes}\:{cos}\pi=−\mathrm{1}\:{typo}.. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
1)(z+αb)^3 =α^3 ×1 (z+αb)^3 =α^3 ×e^(i2kπ) z+αb=α×e^((i2kπ)/3) z_1 +αb=α×e^((i×0×π)/3) =α(cos0^o +isin0^o ) (x_1 +iy_1 )+αb=α(1+i×0) (x_1 ,y_1 )={(α−αb),0}←A z_2 +αb=α×e^((i×2×1×π)/3) =α(cos120^o +isin120^o ) (x_2 +iy_2 )=−αb+α(((−1)/2)+((i×(√3))/2)) (x_2 ,y_2 )={(−αb−(α/2)),((α(√3))/2)}←B z_3 +αb=α×e^((i×2×−1×π)/3) =α(cos120^o −isin120^o ) (x_3 +iy_3 )+αb=α(((−1)/2)−((i(√3))/2)) (x_3 ,y_3 )={(−αb−(α/2)),((−α(√3))/2)}←C BC=α(√3) perpendicular distance from A to BC is α−αb+αb+(α/2)=((3α)/2) so area ABC=(1/2)×α(√3) ×((3α)/2)=((3(√3) α^2 )/4)=1.30α^2
$$\left.\mathrm{1}\right)\left({z}+\alpha{b}\right)^{\mathrm{3}} =\alpha^{\mathrm{3}} ×\mathrm{1} \\ $$$$\left({z}+\alpha{b}\right)^{\mathrm{3}} =\alpha^{\mathrm{3}} ×{e}^{{i}\mathrm{2}{k}\pi} \\ $$$${z}+\alpha{b}=\alpha×{e}^{\frac{{i}\mathrm{2}{k}\pi}{\mathrm{3}}} \\ $$$${z}_{\mathrm{1}} +\alpha{b}=\alpha×{e}^{\frac{{i}×\mathrm{0}×\pi}{\mathrm{3}}} =\alpha\left({cos}\mathrm{0}^{{o}} +{isin}\mathrm{0}^{{o}} \right) \\ $$$$\left({x}_{\mathrm{1}} +{iy}_{\mathrm{1}} \right)+\alpha{b}=\alpha\left(\mathrm{1}+{i}×\mathrm{0}\right) \\ $$$$\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)=\left\{\left(\alpha−\alpha{b}\right),\mathrm{0}\right\}\leftarrow{A} \\ $$$$ \\ $$$${z}_{\mathrm{2}} +\alpha{b}=\alpha×{e}^{\frac{{i}×\mathrm{2}×\mathrm{1}×\pi}{\mathrm{3}}} =\alpha\left({cos}\mathrm{120}^{{o}} +{isin}\mathrm{120}^{{o}} \right) \\ $$$$\left({x}_{\mathrm{2}} +{iy}_{\mathrm{2}} \right)=−\alpha{b}+\alpha\left(\frac{−\mathrm{1}}{\mathrm{2}}+\frac{{i}×\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right)=\left\{\left(−\alpha{b}−\frac{\alpha}{\mathrm{2}}\right),\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}\leftarrow{B} \\ $$$${z}_{\mathrm{3}} +\alpha{b}=\alpha×{e}^{\frac{{i}×\mathrm{2}×−\mathrm{1}×\pi}{\mathrm{3}}} =\alpha\left({cos}\mathrm{120}^{{o}} −{isin}\mathrm{120}^{{o}} \right) \\ $$$$\left({x}_{\mathrm{3}} +{iy}_{\mathrm{3}} \right)+\alpha{b}=\alpha\left(\frac{−\mathrm{1}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\left({x}_{\mathrm{3}} ,{y}_{\mathrm{3}} \right)=\left\{\left(−\alpha{b}−\frac{\alpha}{\mathrm{2}}\right),\frac{−\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}\leftarrow{C} \\ $$$${BC}=\alpha\sqrt{\mathrm{3}}\: \\ $$$${perpendicular}\:{distance}\:{from}\:{A}\:{to}\:{BC}\:{is} \\ $$$$\alpha−\alpha{b}+\alpha{b}+\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{3}\alpha}{\mathrm{2}} \\ $$$${so}\:{area}\:{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×\alpha\sqrt{\mathrm{3}}\:×\frac{\mathrm{3}\alpha}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}\:\alpha^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}.\mathrm{30}\alpha^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by rahul 19 last updated on 05/Dec/18
thank you sir!���� colourful solñ!
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
your posts are unique...
$${your}\:{posts}\:{are}\:{unique}… \\ $$

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