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Question Number 51600 by malwaan last updated on 28/Dec/18

solve (sinθ)Z^2 −i(cosθ)Z+(1/4) sinθ=0

$$\mathrm{solve} \\ $$$$\left(\mathrm{sin}\theta\right)\mathrm{Z}^{\mathrm{2}} −\mathrm{i}\left(\mathrm{cos}\theta\right)\mathrm{Z}+\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{sin}\theta=\mathrm{0} \\ $$

Commented by Abdo msup. last updated on 29/Dec/18

Δ=(−icosθ)^2 −4sinθ((1/4)sinθ)=−cos^2 θ −sin^2 θ =−1 ⇒ Z_1 = ((icosθ+i)/(2sinθ)) =i ((2cos^2 ((θ/2)))/(4sin((θ/2))cos((θ/2)))) = (i/(2tan((θ/2)))) Z_2 =((icosθ−i)/(2sinθ)) =−i((1−cosθ)/(2sinθ)) =−i((2sin^2 ((θ/2)))/(4sin((θ/2))cos((θ/2)))) =−(i/2)tan((θ/2)) (we suppose sinθ≠0) if sinθ=0 ⇒cosθ =+^− 1 and (e) ⇔−i cosθ Z=0 ⇒ Z=0 .

$$\Delta=\left(−{icos}\theta\right)^{\mathrm{2}} −\mathrm{4}{sin}\theta\left(\frac{\mathrm{1}}{\mathrm{4}}{sin}\theta\right)=−{cos}^{\mathrm{2}} \theta\:−{sin}^{\mathrm{2}} \theta \\ $$$$=−\mathrm{1}\:\Rightarrow\:{Z}_{\mathrm{1}} =\:\frac{{icos}\theta+{i}}{\mathrm{2}{sin}\theta}\:={i}\:\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{4}{sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$=\:\frac{{i}}{\mathrm{2}{tan}\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$${Z}_{\mathrm{2}} =\frac{{icos}\theta−{i}}{\mathrm{2}{sin}\theta}\:=−{i}\frac{\mathrm{1}−{cos}\theta}{\mathrm{2}{sin}\theta}\:=−{i}\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{4}{sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$=−\frac{{i}}{\mathrm{2}}{tan}\left(\frac{\theta}{\mathrm{2}}\right)\:\left({we}\:{suppose}\:{sin}\theta\neq\mathrm{0}\right) \\ $$$${if}\:{sin}\theta=\mathrm{0}\:\:\Rightarrow{cos}\theta\:=\overset{−} {+}\mathrm{1}\:\:\:{and}\:\left({e}\right)\:\Leftrightarrow−{i}\:{cos}\theta\:{Z}=\mathrm{0}\:\Rightarrow \\ $$$${Z}=\mathrm{0}\:. \\ $$

Answered by Tawa1 last updated on 29/Dec/18

Commented by malwaan last updated on 29/Dec/18

thank you please solve it with this method Z=x+iy

$$\mathrm{thank}\:\mathrm{you} \\ $$$$\mathrm{please}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{with}\:\mathrm{this}\:\mathrm{method} \\ $$$$\mathrm{Z}=\mathrm{x}+\mathrm{iy}\:\:\:\: \\ $$

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