Question Number 88678 by Cheyboy last updated on 12/Apr/20 | ||
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$$\boldsymbol{\mathrm{F}}{ind}\:\:\:\sqrt{\boldsymbol{{i}}}+\sqrt{−\boldsymbol{\mathrm{i}}} \\ $$ | ||
Commented by mr W last updated on 12/Apr/20 | ||
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$${both}\:{definitions}\:{are}\:{used}\:\left({in}\:{different}\right. \\ $$$$\left.{countries}\right): \\ $$$$−\pi<{Arg}\left({z}\right)\leqslant\pi\:{or} \\ $$$$\mathrm{0}\leqslant{Arg}\left({z}\right)<\mathrm{2}\pi \\ $$ | ||
Commented by mr W last updated on 12/Apr/20 | ||
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$$\sqrt{{i}}+\sqrt{−{i}} \\ $$$$=\left({e}^{\frac{\pi{i}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\left({e}^{\frac{\mathrm{3}\pi{i}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$={e}^{\frac{\pi{i}}{\mathrm{4}}} +{e}^{\frac{\mathrm{3}\pi{i}}{\mathrm{4}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i} \\ $$$$=\sqrt{\mathrm{2}}{i} \\ $$ | ||
Commented by Cheyboy last updated on 12/Apr/20 | ||
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$${Thank}\:{your}\:{sir} \\ $$ | ||
Commented by Tony Lin last updated on 12/Apr/20 | ||
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$${i}={cos}\frac{\pi}{\mathrm{2}}+{isin}\frac{\pi}{\mathrm{2}} \\ $$$$\sqrt{{i}}={cos}\frac{\pi}{\mathrm{4}}+{isin}\frac{\pi}{\mathrm{4}} \\ $$$$−{i}={cos}\left(−\frac{\pi}{\mathrm{2}}\right)+{isin}\left(−\frac{\pi}{\mathrm{2}}\right) \\ $$$$\sqrt{−{i}}={cos}\left(−\frac{\pi}{\mathrm{4}}\right)+{isin}\left(−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\sqrt{{i}}+\sqrt{−{i}}={cos}\frac{\pi}{\mathrm{4}}+{cos}\left(−\frac{\pi}{\mathrm{4}}\right)+{i}\left[{sin}\frac{\pi}{\mathrm{4}}+{sin}\left(−\frac{\pi}{\mathrm{4}}\right)\right] \\ $$$$=\sqrt{\mathrm{2}} \\ $$ | ||
Commented by ajfour last updated on 12/Apr/20 | ||
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$$−\pi<{Arg}\left({z}\right)\leqslant\pi \\ $$$${is}\:{this}\:{true},\:{sir}? \\ $$ | ||
Commented by mr W last updated on 12/Apr/20 | ||
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Commented by ajfour last updated on 12/Apr/20 | ||
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$${thanks}\:{Sir}. \\ $$ | ||
Commented by mathmax by abdo last updated on 12/Apr/20 | ||
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$$\sqrt{{i}}+\sqrt{−{i}}=\left({e}^{\frac{{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\left({e}^{−\frac{{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ={e}^{\frac{{i}\pi}{\mathrm{4}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)=\mathrm{2}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}} \\ $$ | ||