What is the Locus of Re[z^2]>1?

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SUMMARY

The locus of Re[z^2]>1 can be determined by substituting z=x+iy, leading to the inequality x^2-y^2>1. This inequality represents the region outside the hyperbola defined by the equation x^2-y^2=1. By testing points such as (0,0) and (2,0), it becomes evident that the area satisfying the inequality is where x^2-y^2 exceeds 1, confirming the hyperbolic nature of the graph.

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Homework Statement


Find the locus of [tex]Re[z^2]>1[/tex]


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The Attempt at a Solution


Subbing in [tex]z=x+iy[/tex] gives [tex]x^2-y^2>1[/tex], but where do I go from there on. It shouldn't be very tough (i.e. including analysis of functions) because the other examples in the same problem are easy.
 
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Well can you graph [tex]x^2-y^2=1[/tex]? It should be clear where it is more than 1 either by intuition or testing some easy points such as (0,0) versus (2,0).
It is the graph of a hyperbola by the way.
 

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