If $|z-i| \leq 4$, find the maximum value of $|iz+13-5i|$
It's easy, the maximum value is $17$.
But, I'm more concerned with the minimum value,
$$|iz+13-5i| \geq |z+i|-|12-5i|$$
Least value of $|z+i|$ can be $0$ and $|12-5i| = 13$
It means the least value of $|iz+13-5i|$ is $-13$? How is this possible?
On
One way to solve both the maximum and the minimum problem is to consider what the various inequalities mean in the complex plane:
$$|z-i| \le 4$$
describes a disk of radius $4$ around $i$. That means it contains exactly the complex values $z$ which have a distance of $4$ or less from $i$.
If you now consider the value $iz$, that's $z$ multiplied by $i$. Multiplication by a fixed complex number means a rotation around the origin, with the argument of the factor as rotation angle, and then a stretching from the origin, with the absolute value of the factor as stretch factor.
In this case, as the factor is $i$, the absolute value is $1$, the whole operation is just a rotation. To find the center of the rotated disc, we just multiply the center of the original disc with the factor: $i\times i=-1$, so the rotated disc is around the center point $-1$, and of course it still has radius $4$.
That means the set if points $iz$ is the disc around $-1$ with radius 4. If you want to consider
$$|iz+13-5i| = |iz-(-13+5i)|,$$
that's the distance from the point $iz$ to the point $-13+5i$.
$-13+5i$ is a fixed point, and as we saw above $iz$ is a disc of radius $4$ around the point $-1$. So what's the minimum/maximum distance from a fixed point to a disc?
That's easy, you connect the fixed point to the center of the disc. To find the minimum, you then go back along that line segment to the point where the segement crosses into the disc, that's your point of smallest distance.
To find the maximum, you extend the line segment until you find another crossing out of the disc, that would be the point of maximum distance.
Since we don't need to find the exact points, we can more easily find the distance between $-13+5i$ and $-1$, the center of the circle that represents $iz$. That distance is
$$|-1-(-13+5i)|= |12 - 5i| = \sqrt{12^2+5^2}=\sqrt{169}=13.$$
Since the points of minimal and maximal distance are $4$ away from that center, on a circle, it means the maximum distance is $13+4=17$ (as you wrote) and the minimum distance is $13-4=9$.
We have $5\geq 4$, but the smallest value that $5$ can take clearly isn't $4$. The smallest value that $5$ can take is $5$.
Just because your triangle inequality coupled with a lower bound on each term gives you a lower bound of $-13$ for $|iz+13-5i|$, that doesn't mean that $-13$ must be possible to reach.