Basic Complex inequalities

Question

It is given that $|z+2+3i| \leq 1$. Show that $3 \leq |2z+1+2i| \leq 7$. I really do not have a clue how to start on this but have proceeded by noting that

$$(2z+1+2i) = z+2+3i + z-1+i$$ and so $$|2z+1+2i| \leq |z+2+3i| + |z-1+i|$$

But this doesn't get me too far. Can anyone help me complete this proof please ?

Answer 1

We have that $$ |2z+1+2i|=2\left|z+\frac{1}{2}+i\right|=2\left|z+\frac{1}{2}+i\pm(\frac{3}{2}+2i)\right|\\=2\left|(z+2+3i)+\left(-\frac{3}{2}-2i\right)\right|.$$ Note that $|-\frac{3}{2}-2i|=\frac{5}{2}$. Now recall that $||u|-|w||\leq |u+w|\leq |u|+|w|$.

Hence if $|z+2+3i|\leq 1$, then $$3=2\left(\frac{5}{2}-1\right)\leq |2z+1+2i|\leq 2\left(1+\frac{5}{2}\right)=7.$$

Answer 2

Let $z'=z+2+3i.$ So $|z'|\leq 1.$ Now $2z+1+2i=2z'-3-4i.$

We have $$ |2z'-3-4i|\leq |2z'|+|-3-4i|=2|z'|+5\leq 2+5=7.$$ We have $$ |2z'-3-4i|=|3+4i-2z'|\geq |3+4i|-|2z'|=5-2|z'|\geq 5-2=3.$$