Complex number inequality plot - modulus of a fraction I have a question. Consider the following:
$$|\frac{z-2}{z+1-i}|\ge1$$
Here's what I did:
$$\frac{|z-2|}{|z+1-i|}\ge1$$
$$ \frac{\sqrt{(x-2)^2+y^2}}{\sqrt{(x+1)^2+(y-1)^2}} \ge1$$
If we square both sides we get
$$ \frac{(x-2)^2+y^2}{(x+1)^2+(y-1)^2} \ge1$$
If we multiply both sides by the denominator we get
$$(x-2)^2+y^2\ge(x+1)^2+(y-1)^2$$
After which we get that $-6x-2+2y\ge0$
Which I believe is correct. I have two questions:

*

*Am I allowed to consider the expression on the left side as two complex numbers which I calculate the magnitude from separately, (like I did in step 1)? I tried multiplying both the denominator and numerator by $z+1+i$ first hand but it turned out to be complicated for some reason.


*Was I correct to multiply both sides by the denominator? Am I only allowed to do that if I am certain the denominator can't be zero? I'm asking because in a different example I had
$$\frac{2(y-x)}{x^2+y^2-2}\le1$$ and the workbook solution was to subtract 1 from both sides - they didn't multiply by the denominator.
 A: *

*Yes you are


*The original inequality has the implicit assumption that the denominator is non-zero (so actually your answer has a small error - it needs to discount the point (-1, 1)) Since this isn't a strict inequality though, multiplying both sides by zero is allowed, the only thing you need to watch out for is if you're multiplying by a negative, in which case you must flip the direction (but we're safe here, since it's a sum of two real squares)
Also more generally - while your solution is valid, it would have been far faster to realise that this is just the set of points lying on the same side as (-1, 1) of the perpendicular bisector of the line through (-1, 1) and (2, 0) - since when equality holds it's just the locus of points equidistant to (-1, 1) and (2, 0).
A: *

*Yes, for each $z\in\Bbb C$, $z-2$ and $z+1-i$ are two complex numbers.

*The expression $\displaystyle\frac{z-2}{z+1-i}$ makes sense if and only if $z+1-i\ne0$. When this occurs, yes, you can multiply bith sides by the denominator.

