# 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:

1. 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.

2. 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.

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