For example, let's take the inequality $0<x^2+2x+4$. The real solutions to the inequality are shown in the graph. If we look at the graph, all real solutions would work for this inequality.
However, how would I express the complex numbers that also answer the question? In other words, what are the real and complex solutions for this inequality.
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An inequation like $z>0$ where $z$ is a complex number is meaningless because you cannot compare complex numbers.
You can modify the question and for instance ask $\Re(z)>0$, which is allowed as the real part of a complex number is... real.
Now let $x:=u+iv$ and let us solve
$$\Re(x^2+2x+4)=\Re(u^2-v^2+2u+4+i(2uv+2v))=(u+1)^2-v^2+2>0.$$
The solution is a portion of the complex plane delimited by an equilaterl hyperbola centered at $(-1,0)$.
But this answers an arbitrary restatement of the problem :-)
Compute $$x^2+2x+4=x^2+2x+1+3=(x+1)^2+3,$$ which is greater than $0$ for any real number $x.$