If $|Z-1-i|=1$ and $|S-1+i|=2$ where $Z$ and $S$ are two complex numbers then find the maximum value of $\frac1{|4Z-1|^2}+\frac1{|3S+1|^2}$
Can we say we need minimum value of $|4Z-1|^2$ and $|3S+1|^2$?
Minimum values of these numbers would lie along normals passing through $(\frac14,0)$ and $(-\frac13,0)$ respectively.
I found the normals and their points of intersection with the circles and got the points $(\frac25,\frac15)$ and $(-\frac35,\frac7{20})$ respectively, and thus got the overall maximum as $1\frac{20}{101}$
Is this correct? (Most likely not)
What's a better way to do this question?
Yes, we need the minimum value of $|4Z-1|^2$ and the minimum value of $|3S+1|^2$, because $$\max_{Z, S}\left(\frac1{|4Z-1|^2}+\frac1{|3S+1|^2}\right)=\frac1{\min_Z|4Z-1|^2}+\frac1{\min_S|3S+1|^2}$$ since, as you must have noticed, the condition on $Z$ and the condition on $S$ are independent.
Since $Z$ is on the circle centered at $1+i$ with radius $1$, the minimal distance from $Z$ to $\frac14$, which is a point outside the circle is the distance from $\frac14$ to the center minus the radius. That is $$\min_Z\left|Z-\frac14\right|=\left|(1+i)-\frac14\right|-1=\frac14.$$
Similarly, $$\min_S\left|S+\frac13\right|=2-\left|(1-i)+\frac13\right|=\frac13.$$
$$\max_{Z, S}\left(\frac1{|4Z-1|^2}+\frac1{|3S+1|^2}\right) =\frac1{\left(\min_Z4|Z-\frac14|\right)^2} +\frac1{\left(\min_S3|S+\frac13|\right)^2}=\frac11+\frac11=2.$$