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Find-i-i-




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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