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Question Number 128083 by AgnibhoMukhopadhyay last updated on 04/Jan/21

((8 − i)/(3 − 2i)) If the expression above is rewritten in the form a + bi, where a and b are real numbers, what is the value of a? A. 2 B. (8/3) C. 3 D. ((11)/3)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{8}\:−\:{i}}{\mathrm{3}\:−\:\mathrm{2}{i}} \\ $$$$\:\mathrm{If}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{above}\:\mathrm{is}\:\mathrm{rewritten}\: \\ $$$$\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:{a}\:+\:{bi},\:\mathrm{where}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are} \\ $$$$\:\mathrm{real}\:\mathrm{numbers},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}? \\ $$$$\:\mathrm{A}.\:\mathrm{2} \\ $$$$\:\mathrm{B}.\:\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\:\mathrm{C}.\:\mathrm{3} \\ $$$$\:\mathrm{D}.\:\frac{\mathrm{11}}{\mathrm{3}} \\ $$

Answered by MJS_new last updated on 04/Jan/21

((α+βi)/(γ+δi))=(((α+βi)(γ−δi))/((γ+δi)(γ−δi)))=((αγ+βδ)/(γ^2 +δ^2 ))+((βγ−αδ)/(γ^2 +δ^2 ))i

$$\frac{\alpha+\beta\mathrm{i}}{\gamma+\delta\mathrm{i}}=\frac{\left(\alpha+\beta\mathrm{i}\right)\left(\gamma−\delta\mathrm{i}\right)}{\left(\gamma+\delta\mathrm{i}\right)\left(\gamma−\delta\mathrm{i}\right)}=\frac{\alpha\gamma+\beta\delta}{\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} }+\frac{\beta\gamma−\alpha\delta}{\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} }\mathrm{i} \\ $$

Answered by A8;15: last updated on 04/Jan/21

((8−i)/(3−2i))=(((8−i)∙(3+2i))/((3−2i)∙(3+2i)))=((24−3i+16i−2i^2 )/(3^2 −(2i)^2 ))= =((13i+24−2∙((√(−1)))^2 )/(9−4∙((√(−1)))^2 ))=((13i+24−2∙(−1))/(9−4∙(−1)))= =((13i+26)/(13))=((13i)/(13))+((26)/(13))=2+i a+bi=2+i { ((a=2)),((b=1)) :}

$$\frac{\mathrm{8}−\mathrm{i}}{\mathrm{3}−\mathrm{2i}}=\frac{\left(\mathrm{8}−\mathrm{i}\right)\centerdot\left(\mathrm{3}+\mathrm{2i}\right)}{\left(\mathrm{3}−\mathrm{2i}\right)\centerdot\left(\mathrm{3}+\mathrm{2i}\right)}=\frac{\mathrm{24}−\mathrm{3i}+\mathrm{16i}−\mathrm{2i}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} −\left(\mathrm{2i}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{13i}+\mathrm{24}−\mathrm{2}\centerdot\left(\sqrt{−\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{9}−\mathrm{4}\centerdot\left(\sqrt{−\mathrm{1}}\right)^{\mathrm{2}} }=\frac{\mathrm{13i}+\mathrm{24}−\mathrm{2}\centerdot\left(−\mathrm{1}\right)}{\mathrm{9}−\mathrm{4}\centerdot\left(−\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{13i}+\mathrm{26}}{\mathrm{13}}=\frac{\mathrm{13i}}{\mathrm{13}}+\frac{\mathrm{26}}{\mathrm{13}}=\mathrm{2}+\mathrm{i} \\ $$$$\mathrm{a}+\mathrm{bi}=\mathrm{2}+\mathrm{i} \\ $$$$\begin{cases}{\mathrm{a}=\mathrm{2}}\\{\mathrm{b}=\mathrm{1}}\end{cases} \\ $$

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