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Find the value of $\sqrt i.$

$\begin{array}{1 1}(A) \; \pm\large\frac{1}{\sqrt 2}(1+i)\\(B)\;\pm\large\frac{1}{\sqrt 2}(1-i)\\(C)\;\pm(1+i)\\(D)\;\pm(1-i)\end{array}$

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  • $(a+b)^2=(a-b)^2+4ab$
Let $\sqrt i=x+iy$
$\Rightarrow\:i=(x+iy)^2=x^2-y^2+2xyi$
Comparing real and imaginary terms on both the sides
$x^2-y^2=0\:\:and\:\:2xy=1$
$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2$
$\Rightarrow\:x^2+y^2=\sqrt {0+1}=1$
$(x^2+y^2)+(x^2-y^2)=1+0=1\:\:and$
$(x^2+y^2)-(x^2-y^2)=1-0=1$
$\Rightarrow\:2x^2=1\:\:2y^2=1$
$\Rightarrow\:x=y=\pm\large\frac{1}{\sqrt 2}$
$\Rightarrow\:\sqrt i=\pm\large\frac{1}{\sqrt 2}$$(1+i)$
answered Jul 11, 2013 by rvidyagovindarajan_1
 

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