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work out the $\sqrt{2i-1}?$

$2i-1=(a+bi)^2$

$a^2+2abi-b^2$

$a^2-b^2=-1$

$2ab=2$


$a^2=b^{-2}$

$b^{-2}-b^2=-1$

$-b^{4}+1=-1$

$b^4=2$

$b=\sqrt[4]{2}$

Can we solve $\sqrt{2i-1}$ in another way?