I am reading Beardon's *Algebra and Geometry*.

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> Show that $z^2=2i$ iff $z=\pm(1+i)$.
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For the problem in question, first I made the multiplication $(1+i)\times(1+i)$ which showed the result but I guess it lacked generality then I proceded by trying to find all the $x$'s and $y$'s with the following procedure, (although I'm not really sure it will solve my problem): multiplying $(x+iy)\times (x+iy)$ which yielded $x^2+2xiy-y^2=2i$. My guess is that solving for $x$ and $y$ is going to prove that but I don't know how to proceed from here, solving for $y$ yielded:

$y\to i (x-(1+i))$

$y\to i (x+(1+i))$

And solving for $x$ yielded:

$x\to -i (y-(1-i))$

$x\to -i (y+(1-i))$

From here, what should I do to show that $z^2=2i$ iff $z=\pm(1+i)$? (assuming my reasoning is valid)