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

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The locus of Re[z^2]>1 can be derived by substituting z=x+iy, leading to the inequality x^2-y^2>1. This represents the region outside the hyperbola defined by the equation x^2-y^2=1. Testing points like (0,0) and (2,0) confirms the behavior of the graph. The discussion emphasizes that understanding the graph's shape is key to solving the problem. Ultimately, the solution involves recognizing the hyperbolic nature of the inequality.
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Homework Statement


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


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The Attempt at a Solution


Subbing in z=x+iy gives x^2-y^2>1, 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 x^2-y^2=1? 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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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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