Can you give me an example?
Prove algebraically that for any complex number $w?0$ the equation $z^2=w$ has exactly two solutions. When I sub in $x+iy$ for $w$, I get $z^2 = x + iy$. Knowing $z$ is complex, then I sub in $x+iy$ for that as well to get $x^2+2xiy+(iy)^2 = x +...
Let $z=x+i?y$ , where $y>0$ . Choose a positive real number $t$ so that $t^2=x^2+y^2$ and pick positive numbers $u$ and $v$ satisfying $u^2=(t+x)/2 , v^2=(t-x)/2$ . Show that $x+i?y=(u+i?v)^2$ . Can you do a similar procedure of finding a square root...