SOLUTION: (a+bi)^2=i determine a and b must be real numbers

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Question 1074213: (a+bi)^2=i
determine a and b
must be real numbers
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%28a%2Bbi%29%5E2=i+

%28a%2Bbi%29%28a%2Bbi%29=i+

a%5E2%2B2abi%2Bb%5E2i%5E2=i

Since i2 = -1

a%5E2%2B2abi%2Bb%5E2%28-1%29=i

a%5E2%2B2abi-b%5E2=i

Set the real part of the left side equal to the real
part of the right side:

and

Set the imaginary part of the left side equal to the 
imaginary part of the right side:
 
a%5E2-b%5E2=0  and  2abi=i
                      b=i%2F%282ai%29
                      b=1%2F%282a%29

Substitute for b in

a%5E2-b%5E2=0
a%5E2-%281%2F%282a%29%29%5E2=0
a%5E2-1%2F%284a%5E2%29=0
4a%5E4-1=0
%282a%5E2-1%29%282a%5E2%2B1%29=0

2a%5E2-1=0; 2a%5E2%2B1=01
2a%5E2=1;   2a%5E2=-1
a%5E2=1%2F2;  a%5E2=-1%2F2
a=+%22%22+%2B-+sqrt%281%2F2%29%3B+++%7B%7B%7Ba%5E2=%22%22+%2B-+sqrt%28-1%2F2%29
a=+%22%22+%2B-+1%2Fsqrt%282%29; a=%22%22+%2B-i%2Asqrt%281%2F2%29
a=%22%22+%2B-+sqrt%282%29%2F2; ignore the imaginary value for a
Substitute in:
b=1%2F%282a%29
b=1%2F%282%28%22%22+%2B-+sqrt%282%29%2F2%29%29
b=1%2F%28cross%282%29%28%22%22+%2B-+sqrt%282%29%2Fcross%282%29%29%29 
b=1%2F%28%22%22+%2B-+sqrt%282%29%29 
b=%22%22+%2B-+sqrt%282%29%2F2

So there are two solutions:

a=%22%22+%2B-+sqrt%282%29%2F2; b=%22%22+%2B-+sqrt%282%29%2F2

One solution is where a and b are both positive,
and the other solution is when they are both
negative.

Edwin