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Solving inequalities with complex numbers without breaking a sweat

Let us take a look at a problem that came in JEE Mains 2018 CBT.

If $|z-3+2i|\le 4$ then the difference between the greatest value and the least value of $|z|$ is:
1. $4+\sqrt{13}$
2. $2\sqrt{13}$
3. $8$
4. $\sqrt{13}$

 This can be approached using the conventional formula to get the answer. But what if you are not sure about the formula? Even if you do remember the formula, what if you want to arrive at the solution faster?

There is an excellent graphical method to solve this problem.

Consider $z=x+3+(y-2)i$. Note that I am considering this way so that the inequality is neat and clean: $$\sqrt{x^2+y^2}\le 4\\ \therefore x^2+y^2\le 4^2$$
The red region here represents the inequality shown. The region thus contains all and the only values that $x$ and $y$ can take. Let's find out $|z|$ now. $$|z|=\sqrt{(x+3)^2+(y-2)^2}$$ In a graphical point of view, $|z|$ represents the distance between any point in the red area and $(-3,2)$.
It is obvious that the point $(-3,2)$ itself lies inside the red region i.e., satisfy the inequality. Thus the minimum value of $|z|$ must be $0$.
$$|z|_{\text{min}} = 0$$ The maximum value of $|z|$ will be the distance between $(-3,2)$ and the point that lies the farthest away from $(-3,2)$.


It is obvious from the graph that the distance between $(-3,2)$ and $(x,y)$ will be the maximum, when $(x,y)$ satisfies the inequality.
$$\therefore |z|_{\text{max}} = AB = AO+OB\\ AO=\sqrt{(-3-0)^2+(2-0)^2}=\sqrt{13}$$ $OB$ is the radius of the circle $x^2+y^2=4^2$.
$$OB=4\\ \therefore |z|_{\text{max}} = 4+\sqrt{13}$$ Therefore, the answer will be $|z|_{\text{max}}-|z|_{\text{min}} = 4+\sqrt{13}$
If you practice this method a few times and understand how it works, you will be able to answer this kind of problem in a minute!
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