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

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

Answered by MJS_new last updated on 04/Jan/21

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

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

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

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