Let
a=x+iy,z = mi
Substitute these values in the equation.
(-xm^2+1)+i(-m^2y+m)=0
\rightarrow x=y^2
|a|=1
\rightarrow x^2+y^2=1
\rightarrow x^2+x-1=0
\rightarrow x=\frac{-1\pm \sqrt{5}}{2}
x>0 since y has to be real.
x=\frac{\sqrt{5}-1}{2}\qquad y=\sqrt{\frac{\sqrt{5}-1}{2}}
- Manish Shankar good work :)Upvote·0· Reply ·2013-10-17 22:11:06