## anonymous 5 years ago y^2=-i; where i is a imaginary root. y?

1. anonymous

if you imagine y as $r*(\cos(\theta)+i \sin (\theta))$ then : r=+1 or -1 $\theta$=$-\pi/4$

2. anonymous

Let $y=a+ib$ then $y^2=a^2+2aib-b^2=-i$ hence $a^2-b^2=0, 2ab=-1$ hence $b=-1/(2a)$ $a^4=1/4$ $a=\pm \sqrt{1/2}$ then $y=(1/\sqrt{2})(1-i)$ or $y=(1/\sqrt{2})(-1+i)$

3. anonymous

got it?