Answer with Step-by-step explanation:
let z= x+iy
Thus we have
[tex]z^2=(x+iy)^2\\\\z^{2}=x^2+(iy)^2+2ixy\\\\z^2=x^2-y^2+2ixy\\\\Re(z^2)=x^2-y^2[/tex]
Thus for [tex]Re(z^2)=4\\\\x^{2}-y^2=4\\\\\frac{x^2}{2^2}-\frac{y^2}{2^2}=1.........(i)[/tex]
Comparing the above equation with [tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
The equation i above is equation of a hyperbola.
