Question

if z^2=(0,1) find z​

Answer 1

Please see the attachment

Answer 2

Answer:

$z=x+iy$

$z^2=x^2-y^2+2xyi$

$=(0,1)$

$\Rightarrow x^y-y^2=0\\\Rightarrow x^2 =y^2$

$\Rightarrow x=y$

$2xy=1\\2x^2=1\\x^2=\frac{1}{2}$

$x=\frac{1}{\sqrt{2} }$

$z=x+iy$

$=\frac{1}{\sqrt{2} }+ \frac{1}{\sqrt{2} }i$

#SPJ3