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Question Number 14002 by Joel577 last updated on 26/May/17
If (√(−1)) = i, then   what is the value of (√i) ?
$$\mathrm{If}\:\sqrt{−\mathrm{1}}\:=\:{i},\:\mathrm{then}\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\sqrt{{i}}\:? \\ $$
Answered by ajfour last updated on 26/May/17
i=e^(i((π/2)+2nπ))   (√i)=e^(i((π/4)+nπ))       =cos ((π/4)+nπ)+isin ((π/4)+nπ)      if n is even we have  (√i)=(1/( (√2)))+(i/( (√2)))      if n is odd,  (√i)=−((1/( (√2)))+(i/( (√2))))   so  (√i) =±(((1+i)/( (√2))))  .
$${i}={e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi\right)} \\ $$$$\sqrt{{i}}={e}^{{i}\left(\frac{\pi}{\mathrm{4}}+{n}\pi\right)} \\ $$$$\:\:\:\:=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+{n}\pi\right)+{i}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+{n}\pi\right) \\ $$$$\:\:\:\:{if}\:{n}\:{is}\:{even}\:{we}\:{have} \\ $$$$\sqrt{{i}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:{if}\:{n}\:{is}\:{odd}, \\ $$$$\sqrt{{i}}=−\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\: \\ $$$${so}\:\:\sqrt{{i}}\:=\pm\left(\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\right)\:\:. \\ $$
Commented by tawa tawa last updated on 26/May/17
God bless you sir.....
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}…..\: \\ $$
Answered by RasheedSindhi last updated on 26/May/17
Let (√i)  =x+iy        (x+iy)^2 =i         x^2 +2xyi−y^2 =0+i         x^2 −y^2 =0 ∧ 2xy=1        (x+y)(x−y)=0  ∧  xy=(1/2)    (  x=−y ∣ x=y  ) ∧  xy=1/2     •If  x=y⇒x^2 =1/2⇒x=±(1/( (√2)))        y=±(1/( (√2)))     •If x=−y⇒−x^2 =1/2⇒x=±(i/( (√2)))         y=∓(i/( (√2)))    •  (√i) =x+iy=±(1/( (√2)))+i(±(1/( (√2))))             =±((1+i)/( (√2)))    •(√i) =x+iy=±(i/( (√2)))+i(∓(i/( (√2))))          =±(i/( (√2)))+(∓(i^2 /( (√2))))=±((1+i)/( (√2)))  Hence in both cases        (√i)=±((1+i)/( (√2)))
$$\mathrm{Let}\:\sqrt{\mathrm{i}}\:\:=\mathrm{x}+\mathrm{iy} \\ $$$$\:\:\:\:\:\:\left(\mathrm{x}+\mathrm{iy}\right)^{\mathrm{2}} =\mathrm{i} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{2xyi}−\mathrm{y}^{\mathrm{2}} =\mathrm{0}+\mathrm{i} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} =\mathrm{0}\:\wedge\:\mathrm{2xy}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}−\mathrm{y}\right)=\mathrm{0}\:\:\wedge\:\:\mathrm{xy}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\left(\:\:\mathrm{x}=−\mathrm{y}\:\mid\:\mathrm{x}=\mathrm{y}\:\:\right)\:\wedge\:\:\mathrm{xy}=\mathrm{1}/\mathrm{2} \\ $$$$\:\:\:\bullet\mathrm{If}\:\:\mathrm{x}=\mathrm{y}\Rightarrow\mathrm{x}^{\mathrm{2}} =\mathrm{1}/\mathrm{2}\Rightarrow\mathrm{x}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\mathrm{y}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\bullet\mathrm{If}\:\mathrm{x}=−\mathrm{y}\Rightarrow−\mathrm{x}^{\mathrm{2}} =\mathrm{1}/\mathrm{2}\Rightarrow\mathrm{x}=\pm\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\mathrm{y}=\mp\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\bullet\:\:\sqrt{\mathrm{i}}\:=\mathrm{x}+\mathrm{iy}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{i}\left(\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\pm\frac{\mathrm{1}+\mathrm{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\bullet\sqrt{\mathrm{i}}\:=\mathrm{x}+\mathrm{iy}=\pm\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}+\mathrm{i}\left(\mp\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\pm\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}+\left(\mp\frac{\mathrm{i}^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}}\right)=\pm\frac{\mathrm{1}+\mathrm{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{Hence}\:\mathrm{in}\:\mathrm{both}\:\mathrm{cases} \\ $$$$\:\:\:\:\:\:\sqrt{\mathrm{i}}=\pm\frac{\mathrm{1}+\mathrm{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$
Commented by chux last updated on 26/May/17
wow...... thanks
$$\mathrm{wow}……\:\mathrm{thanks} \\ $$

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