Question 42808: (a+bi)^2=2i
then what are the possible values of a and b?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (a+bi)^2=2i
then what are the possible values of a and b?
a^2+2abi-b^2=0+2i
(a^2-b^2)+2abi = 0+2i
So, a^2-b^2=0 and 2ab=2 or ab=1
Since ab=1, a=1/b
Substitute to get:
(1/b)^2-b^2=0
1/b^2 - b^2=0
1- b^4=0
b^4=1
b=1 or b=-1
Since ab=1, a=1 when b=1 or a=-1 when b=-1
Solution
If a=b=1 then a+bi = 1+i
If a=b=-1 then a+bi = -1-i
Cheers,
Stan H.
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