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Question Number 88678 by Cheyboy last updated on 12/Apr/20

Find (√i)+(√(−i))

$$\boldsymbol{\mathrm{F}}{ind}\:\:\:\sqrt{\boldsymbol{{i}}}+\sqrt{−\boldsymbol{\mathrm{i}}} \\ $$

Commented by mr W last updated on 12/Apr/20

both definitions are used (in different countries): −π<Arg(z)≤π or 0≤Arg(z)<2π

$${both}\:{definitions}\:{are}\:{used}\:\left({in}\:{different}\right. \\ $$$$\left.{countries}\right): \\ $$$$−\pi<{Arg}\left({z}\right)\leqslant\pi\:{or} \\ $$$$\mathrm{0}\leqslant{Arg}\left({z}\right)<\mathrm{2}\pi \\ $$

Commented by mr W last updated on 12/Apr/20

(√i)+(√(−i)) =(e^((πi)/2) )^(1/2) +(e^((3πi)/2) )^(1/2) =e^((πi)/4) +e^((3πi)/4) =((√2)/2)+i((√2)/2)−((√2)/2)+((√2)/2)i =(√2)i

$$\sqrt{{i}}+\sqrt{−{i}} \\ $$$$=\left({e}^{\frac{\pi{i}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\left({e}^{\frac{\mathrm{3}\pi{i}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$={e}^{\frac{\pi{i}}{\mathrm{4}}} +{e}^{\frac{\mathrm{3}\pi{i}}{\mathrm{4}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i} \\ $$$$=\sqrt{\mathrm{2}}{i} \\ $$

Commented by Cheyboy last updated on 12/Apr/20

Thank your sir

$${Thank}\:{your}\:{sir} \\ $$

Commented by Tony Lin last updated on 12/Apr/20

i=cos(π/2)+isin(π/2) (√i)=cos(π/4)+isin(π/4) −i=cos(−(π/2))+isin(−(π/2)) (√(−i))=cos(−(π/4))+isin(−(π/4)) (√i)+(√(−i))=cos(π/4)+cos(−(π/4))+i[sin(π/4)+sin(−(π/4))] =(√2)

$${i}={cos}\frac{\pi}{\mathrm{2}}+{isin}\frac{\pi}{\mathrm{2}} \\ $$$$\sqrt{{i}}={cos}\frac{\pi}{\mathrm{4}}+{isin}\frac{\pi}{\mathrm{4}} \\ $$$$−{i}={cos}\left(−\frac{\pi}{\mathrm{2}}\right)+{isin}\left(−\frac{\pi}{\mathrm{2}}\right) \\ $$$$\sqrt{−{i}}={cos}\left(−\frac{\pi}{\mathrm{4}}\right)+{isin}\left(−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\sqrt{{i}}+\sqrt{−{i}}={cos}\frac{\pi}{\mathrm{4}}+{cos}\left(−\frac{\pi}{\mathrm{4}}\right)+{i}\left[{sin}\frac{\pi}{\mathrm{4}}+{sin}\left(−\frac{\pi}{\mathrm{4}}\right)\right] \\ $$$$=\sqrt{\mathrm{2}} \\ $$

Commented by ajfour last updated on 12/Apr/20

−π<Arg(z)≤π is this true, sir?

$$−\pi<{Arg}\left({z}\right)\leqslant\pi \\ $$$${is}\:{this}\:{true},\:{sir}? \\ $$

Commented by mr W last updated on 12/Apr/20

Commented by ajfour last updated on 12/Apr/20

thanks Sir.

$${thanks}\:{Sir}. \\ $$

Commented by mathmax by abdo last updated on 12/Apr/20

(√i)+(√(−i))=(e^((iπ)/2) )^(1/2) +(e^(−((iπ)/2)) )^(1/2) =e^((iπ)/4) +e^(−((iπ)/4)) =2cos((π/4))=2×(1/(√2))=(√2)

$$\sqrt{{i}}+\sqrt{−{i}}=\left({e}^{\frac{{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\left({e}^{−\frac{{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ={e}^{\frac{{i}\pi}{\mathrm{4}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}\right)=\mathrm{2}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}} \\ $$

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