Question Number 121127 by Study last updated on 05/Nov/20
$$\sqrt{\mathrm{3}+\mathrm{2}{i}}=? \\ $$
Answered by MJS_new last updated on 05/Nov/20
$$\mathrm{3}+\mathrm{2i}=\sqrt{\mathrm{13}}\mathrm{e}^{\mathrm{i}\:\mathrm{arctan}\:\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\sqrt{\mathrm{3}+\mathrm{2i}}=\mathrm{13}^{\mathrm{1}/\mathrm{4}} \mathrm{e}^{\mathrm{i}\:\frac{\mathrm{arctan}\:\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{2}}} \\ $$$$\mathrm{or} \\ $$$$\sqrt{{a}+{b}\mathrm{i}}=\frac{\sqrt{\mathrm{2}\left({a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}}{\mathrm{2}}+\mathrm{sign}\:{b}\:\frac{\sqrt{\mathrm{2}\left(−{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$$\sqrt{\mathrm{3}+\mathrm{2i}}=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{13}}}}{\mathrm{2}}+\frac{\sqrt{−\mathrm{6}+\mathrm{2}\sqrt{\mathrm{13}}}}{\mathrm{2}}\mathrm{i} \\ $$
Answered by Dwaipayan Shikari last updated on 05/Nov/20
$${Generally} \\ $$$$\sqrt{{x}+{i}\sqrt{{y}}}\:=\sqrt{{a}}+{i}\sqrt{{b}} \\ $$$${x}+{i}\sqrt{{y}}\:={a}−{b}+\mathrm{2}{i}\sqrt{{ab}} \\ $$$${x}={a}−{b}\:\:\:\:\:,\:\:\:\:{y}=\mathrm{4}{ab} \\ $$$${a}+{b}=\sqrt{{x}^{\mathrm{2}} +{y}} \\ $$$${a}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{y}}}{\mathrm{2}}\:\:,\:\:{b}=\frac{{x}−\sqrt{{x}^{\mathrm{2}} +{y}}}{\mathrm{2}} \\ $$$$\sqrt{{x}+{i}\sqrt{{y}}}=\sqrt{\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{y}}}{\mathrm{2}}}+{i}\sqrt{\frac{{x}−\sqrt{{x}^{\mathrm{2}} +{y}}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{3}+{i}\sqrt{\mathrm{4}}}=\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}}}+{i}\sqrt{\frac{\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}} \\ $$
Answered by floor(10²Eta[1]) last updated on 05/Nov/20
$$\sqrt{\mathrm{3}+\mathrm{2i}}=\mathrm{a}+\mathrm{bi},\:\mathrm{a},\mathrm{b}\in\mathbb{R} \\ $$$$\mathrm{3}+\mathrm{2i}=\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} +\mathrm{2abi} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} =\mathrm{3} \\ $$$$\mathrm{2ab}=\mathrm{2}\therefore\mathrm{ab}=\mathrm{1} \\ $$$$\mathrm{b}^{\mathrm{4}} +\mathrm{3b}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{b}^{\mathrm{2}} =\frac{−\mathrm{3}\pm\sqrt{\mathrm{13}}}{\mathrm{2}}\Rightarrow\mathrm{b}=\pm\sqrt{\frac{\sqrt{\mathrm{13}}−\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{a}=\pm\sqrt{\frac{\sqrt{\mathrm{13}}+\mathrm{3}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{3}+\mathrm{2i}}=\sqrt{\frac{\sqrt{\mathrm{13}}+\mathrm{3}}{\mathrm{2}}}+\mathrm{i}\sqrt{\frac{\sqrt{\mathrm{13}}−\mathrm{3}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{3}+\mathrm{2i}}=−\sqrt{\frac{\sqrt{\mathrm{13}}+\mathrm{3}}{\mathrm{2}}}−\mathrm{i}\sqrt{\frac{\sqrt{\mathrm{13}}−\mathrm{3}}{\mathrm{2}}} \\ $$