Question 1040725: Prove that √i = (1+i)/√2
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! an imaginary number can be represented by a+bi where a and b are real numbers, then
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(a+bi)^2 = i
:
(a^2 - b^2) + (2ab)i = 0 + 1i
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note that i^2 = -1
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now equate the real and imaginary parts and we have
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1) a^2 - b^2 = 0
2) 2ab = 1
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a^2 - b^2 = 0 means that a = + or - b
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if a = -b then equation 2 becomes -2b^2 = 1, this can not be solve for b a real number
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so we use a = b and equation 2 becomes 2a^2 = 1 and a = b = 1/square root(2) or a = b = -1/square root(2)
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therefore
:
a+bi = (1/square root(2)) + (1/square root(2))i = (1/square root(2))(1+i)
:
a+bi = (-1/square root(2))(1+i)
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note that there are two answers to the square root(i)
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