Find the two square roots for the following complex number.
Write your answers in standard form: 2i
Let the answer be a complex number
a+bi with a,b real. Then
(a+bi)Č = 2i
aČ+2abi+bČiČ = 2i
aČ+2abi+bČ(-1) = 2i
aČ+2abi-bČ=2i
Equate real parts and equate imaginary
parts:
aČ-bČ = 0 2abi = 2i
aČ = bČ ab = 1
So the only possible real solutions
are (a,b) = (1,1) and (a,b) = (-1,-1)
So the two square roots of 2i are
1+i and -1-i <--answers
Checking:
(1+i)Č = 1+2i+iČ = 1+2i+(-1) = 2i
Edwin
