Question- If $z_0,z$ are complex numbers s.t. $|z-i|\leq2 $ & $z_0=5+3i$, find $|iz+z_0|_{max}$
My Approach -
by Triangle Inequality, we know $|z_1+z_2|\leq|z_1|+|z_2|\implies|z_1+z_2|_{max}=|z_1|+|z_2|$
given $z_0=5+3i$, we get
$|z_0+iz|_{max}=|z_0|+|iz|=\sqrt{5^2 + 3^2}+|iz|=\sqrt{34} + |z|$
Now, its given
$\because|z-i|\leq2\implies|z-i|_{max}=|z+(-i)|_{max}=2\implies|z|+|-i|=2\implies|z|=1$
From above, we can write $|z_0+iz|_{max}=\sqrt{34}+|z|=\sqrt{34}+1$
However, my teacher says the correct answer is 7.
Can somebody help me !
If $|z - i| \le 2$, then $z = 3i$ certainly satisfies this inequality, since $|3i - i| = |2i| = 2$, yet $|3i| = 3 > 1$. So you clearly have an error. Where is it? Well, the triangle inequality does say that $$2 = |z-i| = |z+(-i)| \color{red}{\le} |z| + |-i| = |z| + 1. \tag{1}$$ But the problem here is that you write $$|z - i| = 2 \implies |z| + 1 = 2,$$ when instead the inequality $(1)$ clearly is saying $$2 \color{red}{\le} |z| + 1,$$ or $$|z| \color{red}{\ge} 1.$$
A better approach would be to let $w = iz+z_0$ and note that if $|z-i| \le 2$, then $$\begin{align} 2 &\ge |z-i| \\ &= |i||z-i| \\ &= |iz-i^2| \\ &= |iz+1| \\ &= |iz+(5+3i)-(4+3i)| \\ &= |iz+z_0 - (4+3i)| \\ &= |w - (4+3i)|. \tag{2} \end{align}$$ We want to find, for all $w$ satisfying $(2)$, the one with maxmimum $|w|$. Geometrically, what is $(2)$? It is the disk in the complex plane with radius $2$ and center $4+3i$. The point in this disk that is furthest away from the origin, and thus has largest magnitude, is the one that lies on the line joining $0$ and $4+3i$. Since $|4+3i| = 5$ and the disk has radius $2$, this means $|w|_{\text{max}} = 5 + 2 = 7$.