Minimum value of $|z|$ if $|z-(4+3i)|=3$ By triangle inequality, we have that 
$$|z+(-4-3i)| \leq |z|+|4+3i| $$ 
So $$3 \leq |z| +5 $$
That means $$ -2 \leq |z| $$
But geometry says it should be $+2$ . 
What am I doing wrong? How can I prove it without geometry of the locus ?
 A: Geometrically, this means that the distance between $z$ and $4+3i$ is $3$ and $|z|$ is the distance form $z$ to the origin. So, our minimum value would like on the line between the origin and $4 + 3i$ that is $3$ units away form $4+ 3i$. We easily see that this is $5 - 3 = 2$. 
A: Even though you don't want geometry, solving analytically is not so frightening.
You want to find the minimum of $f(x,y)=x^2+y^2$ while $g(x,y)=(x-4)^2+(y-3)^2-9=0$.
To get this this we need $\vec{\operatorname{grad}f}=\begin{pmatrix}2x\\ 2y\end{pmatrix} \propto \vec{\operatorname{grad}g}=\begin{pmatrix}2x-8\\ 2y-6\end{pmatrix}$
This is $\begin{cases} x-4 = ax \\ y-3=ay\end{cases}\implies \displaystyle g(x,y)=(x^2+y^2)a^2-9=\frac{25a^2}{(1-a)^2}-9=0$
Solving for $a$ gives $a=\frac 38,\ -\frac 32$.
$\displaystyle f(x,y)=\frac 9{a^2}=64$ or $4$ so $|z_{max}|=8$ and $|z_{min}|=2$.
A: Using Inequalities Among Complex Numbers, $$|z+w|\le|z|+|w|\implies|z+(-4-3i)|\le|z|+|-4-3i|=|z|+5$$ 
So, we need $3\le|z|+5\iff|z|\ge3-5=-2$ which is always true
and $$||z|-|w||\le|z-w|\implies||z|-5|\le|z-(4+3i)|$$
If $|z|\ge5,3\ge|z|-5\iff|z|\le3+5$
Else $|z|<5\implies5-|z|\le3\iff|z|\ge5-3$
