Find all the complex numbers such that $\displaystyle z^2=1-i$.

My attempt : I wrote $\displaystyle z=(a+ib)$ and with a little bit of some algebra I reached that $\displaystyle ab=-\frac{1}{2}$ which mean that if $\displaystyle ab=-\frac{1}{2}$ then $\displaystyle z^2=1-i$ is satisfied. So the solution is all z such that $\displaystyle \forall$ a,b $\displaystyle \in \mathbb{R}$ / $\displaystyle ab=-\frac{1}{2}$ with $\displaystyle z=a+ib$. I'm sure there is a nicer way to write the answer. Can you check out my answer and write it in a better way if it is right? Thanks in advance.