Question Number 167911 by MathsFan last updated on 29/Mar/22 | ||
$${If}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1} \\ $$$$\:{then},\:\:\frac{\mathrm{1}+{b}+{ia}}{\mathrm{1}+{b}−{ia}}=? \\ $$ | ||
Answered by MJS_new last updated on 29/Mar/22 | ||
$$\frac{{b}+\mathrm{1}+{a}\mathrm{i}}{{b}+\mathrm{1}−{a}\mathrm{i}}=\frac{\left({b}+\mathrm{1}+{a}\mathrm{i}\right)^{\mathrm{2}} }{\left({b}+\mathrm{1}−{a}\mathrm{i}\right)\left({b}+\mathrm{1}+{a}\mathrm{i}\right)}= \\ $$$$=\frac{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{1}+\mathrm{2}{a}\left({b}+\mathrm{1}\right)\mathrm{i}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{1}}= \\ $$$$\:\:\:\:\:\left[{b}^{\mathrm{2}} =\mathrm{1}−{a}^{\mathrm{2}} \right] \\ $$$$=\frac{−\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{2}+\mathrm{2}{a}\left({b}+\mathrm{1}\right)\mathrm{i}}{\mathrm{2}{b}+\mathrm{2}}= \\ $$$$=\frac{−{a}^{\mathrm{2}} +{b}+\mathrm{1}}{{b}+\mathrm{1}}+{a}\mathrm{i}= \\ $$$$\:\:\:\:\:\left[{a}^{\mathrm{2}} =\mathrm{1}−{b}^{\mathrm{2}} \right] \\ $$$$=\frac{{b}^{\mathrm{2}} +{b}}{{b}+\mathrm{1}}+{a}\mathrm{i}= \\ $$$$={b}+{a}\mathrm{i} \\ $$ | ||
Commented by MathsFan last updated on 29/Mar/22 | ||
$${thanks} \\ $$ | ||