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?
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?