Why does the triangle inequality not work here? Question:

If $\left|z-i\right|\le2$  and $z_{1}=5+3i$, then find the maximum value of $\left|iz+z_{1}\right|$

My Approach:

$\left|iz+z_{1}\right|\le\left|iz\right|+\left|z_{1}\right|$.
We have, $\left|iz\right|=\left|i\right|\cdot\left|z\right|=\left|z\right|$ and $\left|z_{1}\right|=\sqrt{5^{2}+3^{2}}=\sqrt{34}$
Thus, $\left|iz+z_{1}\right|_{max}=\sqrt{34}+\left|z\right|_{max}=\sqrt{34}+3$

($\left|z\right|_{max}=3$, I got this using the graph of  $\left|z-i\right|\le2$)
But this is not the answer, the answer given is $7$. I think I'm getting the wrong answer because of the restriction $\left|z-i\right|\le2$ on $z$, but I'm not able to proceed. Please help.
 A: The way you solved it the triangle inequality still holds.  That is it is true that $\sqrt {34} + 3 \ge 7$ but you are not using all of the information most efficiently.
I would suggest you work this out geometrically.  Being able to make these sketches will develop your intuition.
$|z-i|\le 2$ is a disk centered at $i.$
Then you have to work out the distance from  $iz_0$ to the farthest point on the disc.
But here is the algebraic method.
$|iz + 5 + 3i| = |i(z-i) + 4 + 3i| \le |(z-i)| + |4+3i| = 2 + 5$
A: 
The disk $|z - i| ≤ 2$ can be represented as $x^2 + (y - 1)^2 ≤ 4$. Now we can apply the transformation $z \mapsto iz + z_1$ on this disk.
First, multiplying by $i$ results in a counterclockwise rotation of $\pi/2$ around the origin. Using the fact that $\frac{y}{x} \frac{-x}{y} = -1$ to be perpendicular, $(x, y)$ rotated by $\pi/2$ gives $(-y, x)$, which has the same distance as $(x,y)$ to the origin. Thus applying the transformation $x \mapsto -(-y), y \mapsto -(x)$ to the circle, we obtain the equation of the rotated circle $(y)^2 + (-x - 1)^2 = (x +1)^2 + y^2 = 4$.
The translation $iz \mapsto iz + z_1$ is straightforward. Translating the rotated circle by $5 \choose 3$ units gives $(x - 4)^2 + (y - 3)^2 = 4$.
And now, the modulus represents the distance to the origin. The maximum distance occurs when the component in the direction ${4 \choose 3}$, the centre, is maximised. Since the modulus of $4 + 3i$ is $5$, we can achieve a maximum of $2$ units in this direction.
Hence the maximum value of $|iz + z_1|$ is $5 + 2 = \boxed{7}$.
