Let A be a complex number and B a real number. Show that the equation
|z|^2+Re(Az)+B=0 has a solution if and only if |A|^2 >= 4B. If this is so, show that the solutions set is a circle or a single point.
Since A is a complex number, we can suppose A=a+bi, where a,b are real numbers. B is a real number. We can also suppose z=x+iy, where x and y are real numbers.
From the equation ...
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