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Question

how would I prove that the square root of 2 plus the square root of -2 all divided by 2 is the square root of i? Thanks so much!

Answer 1

{SQR2 + SQR[-2]}/2 = [SQR2 + i SQR2]/2 =

= [1 + i]/SQR2

{[1 + i]/SQR2}^2 = [1 + 2i -1]/2 = 2i/2 = i

So it is the square root of i

Answer 2

Let's first translate the sentence into an expression in math...

(sqrt2 + sqrt(-2)) / 2

= (sqrt2 + sqrt(-1)(2)) / 2

You should know that the sqrt(-1) is i, so we can take the -1 out of the second sqrt and put an i out front

= (sqrt2 + isqrt(2)) / 2

Now, let's factor sqrt(2) out of the numerator since it's common in both terms

= sqrt2(1 + i) / 2

That's the answer; it's not sqrt(i).

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