range of $2$ variable function

Question

If $|z-1|+|z+3|\leq 8$. Then range of $|z-4|$ is, where $z$ is a complex number.

Try: let $z=x+iy$, Then $\sqrt{(x-1)^2+y^2}+\sqrt{(x+3)^2+y^2}\leq 8$

It represent an ellipse whose one focus is at $A(-3,0)$ and $B(1,0)$

And $P(x,y)$ represent a variable point on that ellipse and its interior parts.

so $PA+PB\leq 8$ and length of semimazor axis is $a=4$ and $$ae=2\Rightarrow e=\frac{1}{2}$$

So $b^2=a^2(1-e^2)=12$

Could some help me to solve it, thanks

Or can we solve it without geometry means using triangle inequality.

Answer 1

In order to find the range of $|z-4|$ we need to shift the ellipse $$|z-1|+|z+3|\leq 8$$

$ 4$ units to the left. After the shift we get another ellipse $$|z+3|+|z+7|\leq 8$$ Considering the vertices of the new ellipse we get $$ 1\le |z-4| \le 9 $$

Answer 2

We solve this problem geometrically. $|z-1|+|z+3|\le 8$ describes inside and on the perimeter of an ellipse whose focuses are on $(1,0)$ and $(-3,0)$ and the large diameter is between $(-5,0)$ and $(3,0)$. Therefore the maximal value of $|z-4|$ is $9$ taking place in $(-5,0)$