By the quadratic formula you should get
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From here you need to use that the square root is multivalued.
It is simplest to take roots by using DeMoivre's formula: The nth roots of complex number , in polar form, are given by where k= 0 to n-1 give the n roots. (r must be positive of course, and is the positive real n th root.)
In particular, if [itex]a+ bi= r e^{i\theta}= r(cos(\theta)+ i sin(\theta))[/itex] then the square root of a+ bi is given by and
An alternative approach to the excellent approaches already suggested is to substitute z = a + ib, expand and equate the real and imaginary parts to zero. This gives two simultaneous equations to be solved for a and b, leading to the two solutions z = 2 + i and z = 1 - i.