Question 1196798
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If x is a square root of y, then {{{x = sqrt(y)}}} which rearranges into {{{x^2 = y}}}


Example: 7 is a square root of 49, so {{{7=sqrt(49)}}} which turns into {{{7^2 = 49}}} after squaring both sides.


The claim is {{{sqrt(2)/2+expr(sqrt(2)/2)i}}} is a square root of {{{i}}}, where {{{i = sqrt(-1)}}}


A useful rule of complex numbers is this
{{{(a+bi)^2 = (a^2-b^2) + 2abi}}}
I'll leave it to the student to prove this claim.


In this case, {{{a = sqrt(2)/2}}} and {{{b = sqrt(2)/2}}}


So,
{{{(a+bi)^2 = (a^2-b^2) + 2abi}}}


{{{(sqrt(2)/2+expr(sqrt(2)/2)i)^2 = ((sqrt(2)/2)^2-(sqrt(2)/2)^2) + 2*(sqrt(2)/2)*expr(sqrt(2)/2)i}}}


{{{(sqrt(2)/2+expr(sqrt(2)/2)i)^2 = 0 + 1i}}}


{{{(sqrt(2)/2+expr(sqrt(2)/2)i)^2 = i}}}


This verifies that {{{(sqrt(2)/2+expr(sqrt(2)/2)i)^2}}} is indeed a square root of {{{i}}}


In other words, one solution to {{{x^2 = i}}} is {{{x = sqrt(2)/2+expr(sqrt(2)/2)i}}}


The other solution is {{{x = -sqrt(2)/2-expr(sqrt(2)/2)i}}} which I'll leave to the reader to prove. 
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