Respuesta :

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.

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