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Work out the Sqrt of 3+4i
Supposing the square root of 3+4i is a+ib where a and b is real
Then
(a+ib)2=3 + 4i
(a+ib) (a+ib) = 3 + 4i
a(a+ib) + ib(a+ib) = 3 + 4i
a2 + abi + abi – b2 = 3+4i
(a2-b2) +2abi = 3+4i
By equating the real parts from each side of the equation, and the imaginary parts too.
I a2 – b2 =3
II 2ab =4
4/2a =2/a
Substituing into i a2-4/a2 =3
A4 – 4 = 3a2
A4 -3a2 – 4 =0
(a2 – 4)(a2 + 1) = 0
Therefore a2 = 4 or -1
Since we supposed a is to be real, a2 cannot be -1
Solutions are a =2 or a = -2
Substituting back into b=2/a
When a =2 b =1
When a = -2 b = -1
Therefore the sqrt is 2 +i and -2 -i