KeltingMeith's method is the one usually taught for finding square roots of a complex number, without using polar form. Here is an alternative method which I also encourage students to learn, because the algebra is often easier.
Suppose we want to solve \(z^2 = a+bi\) where \(a\) and \(b\) are real. We let \(z = x+yi\), where \(x\) and \(y\) are both real, and hence our equation is \((x+yi)^2=a+bi\).
Expanding out the LHS and equating real and imaginary parts, we get \(x^2 - y^2 = a\) and \(2xy = b\). So far this is the same as the usual method, and the next step would be to solve simultaneously by substituting the second equation into the first.
However, we can generate another useful equation by noting that \(\lvert z^2 \rvert = \lvert z \rvert ^2\) for any complex number \(z\) and hence we have:
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And now, since \(\lvert (x+yi)^2\rvert = \lvert a+bi \rvert \), we have \(x^2 + y^2 = \sqrt{a^2 + b^2}\).
So we now have two equations in \(x^2\) and \(y^2\) that are easily solved by elimination: \(x^2 - y^2 = a\) and \(x^2 + y^2 = \sqrt{a^2 + b^2}\).
Once you have solved for \(x^2\) and \(y^2\), then you find \(x\) and \(y\) by taking square roots. To know which signs of the roots should be taken, use the equation \(2xy = b\): if \(b > 0\), then \(x\) and \(y\) have the same sign; if \(b < 0\), then \(x\) and \(y\) have opposite signs.