SOLUTION: prove that √ 2/2 + √ 2/2i is a square root of i

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Question 1196798: prove that √ 2/2 + √ 2/2i is a square root of i
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

If x is a square root of y, then x+=+sqrt%28y%29 which rearranges into x%5E2+=+y

Example: 7 is a square root of 49, so 7=sqrt%2849%29 which turns into 7%5E2+=+49 after squaring both sides.

The claim is sqrt%282%29%2F2%2Bexpr%28sqrt%282%29%2F2%29i is a square root of i, where i+=+sqrt%28-1%29

A useful rule of complex numbers is this
%28a%2Bbi%29%5E2+=+%28a%5E2-b%5E2%29+%2B+2abi
I'll leave it to the student to prove this claim.

In this case, a+=+sqrt%282%29%2F2 and b+=+sqrt%282%29%2F2

So,
%28a%2Bbi%29%5E2+=+%28a%5E2-b%5E2%29+%2B+2abi



%28sqrt%282%29%2F2%2Bexpr%28sqrt%282%29%2F2%29i%29%5E2+=+0+%2B+1i

%28sqrt%282%29%2F2%2Bexpr%28sqrt%282%29%2F2%29i%29%5E2+=+i

This verifies that %28sqrt%282%29%2F2%2Bexpr%28sqrt%282%29%2F2%29i%29%5E2 is indeed a square root of i

In other words, one solution to x%5E2+=+i is x+=+sqrt%282%29%2F2%2Bexpr%28sqrt%282%29%2F2%29i

The other solution is x+=+-sqrt%282%29%2F2-expr%28sqrt%282%29%2F2%29i which I'll leave to the reader to prove.

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
The square of the modulus of this number,  sqrt%282%29%2F2 + %28sqrt%282%29%2F2%29%2Ai, is

    r%5E2 = %28sqrt%282%29%2F2%29%5E2 + %28sqrt%282%29%2F2%29%5E2 = 2%2F4+%2B+2%2F4 = 1%2F2 + 1%2F2 = 1.


So, the modulus itself is  r = sqrt%281%29 = 1.


The argument of this number, "a", satisfies 

    tan(a) = %28%28sqrt%282%29%2F2%29%29%2F%28%28sqrt%282%29%2F2%29%29 = 1.


So, the argument is  a = pi%2F4 = 45°.


Hence, the square of this number has the modulus 1%5E2 = 1 and the argument 2a = pi%2F2 = 90°.


It means that the square of the given number is the complex imaginary unit "i".

At this point, the proof is complete.