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If-j-1-a-jb-where-a-b-are-real-express-b-in-terms-of-Answer-2-2-2-1-




Question Number 51571 by Tawa1 last updated on 28/Dec/18
If   α − jβ  =  (1/(a − jb))  ,      where   α, β, a, b  are real, express  b  in terms  of  α, β                          Answer:          ((− β)/(α^2  + β^2  − 2α + 1))
$$\mathrm{If}\:\:\:\alpha\:−\:\mathrm{j}\beta\:\:=\:\:\frac{\mathrm{1}}{\mathrm{a}\:−\:\mathrm{jb}}\:\:,\:\:\:\:\:\:\mathrm{where}\:\:\:\alpha,\:\beta,\:\mathrm{a},\:\mathrm{b}\:\:\mathrm{are}\:\mathrm{real},\:\mathrm{express}\:\:\mathrm{b}\:\:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:\:\alpha,\:\beta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Answer}:\:\:\:\:\:\:\:\:\:\:\frac{−\:\beta}{\alpha^{\mathrm{2}} \:+\:\beta^{\mathrm{2}} \:−\:\mathrm{2}\alpha\:+\:\mathrm{1}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Dec/18
a−ib=(1/(α−iβ))  a−ib=((α+iβ)/(α^2 +β^2 ))=((α/(α^2 +β^2 )))+i((β/(α^2 +β^2 )))  b=−((β/(α^2 +β^2 )))  i think so...
$${a}−{ib}=\frac{\mathrm{1}}{\alpha−{i}\beta} \\ $$$${a}−{ib}=\frac{\alpha+{i}\beta}{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }=\left(\frac{\alpha}{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }\right)+{i}\left(\frac{\beta}{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }\right) \\ $$$${b}=−\left(\frac{\beta}{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }\right) \\ $$$${i}\:{think}\:{so}… \\ $$
Commented by Tawa1 last updated on 28/Dec/18
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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