Find the range of $|3-i(z-1)|$

Question

Original question:

If $|z-2 i| \leq \sqrt{2},$ then the maximum value of $|3-i(z-1)|$ is equal to

My attempt:

(1) $|z-2 i| \leq \sqrt{2}$ is a circle, centred at $(0,2)$ and radius $\sqrt 2$ and $|3-i(z-1)|=|z-(1-3i)|$(multiplying and dividing by $i$) represents the distance of the points lying on the circle from $(1,-3)$.

For minimum distance, it will be the points closest to itself, lying on the line joining the centre and that point i.e. distance between the points minus the radius= $\sqrt{ (0-1)^2+(2-(-3))^2}-\sqrt{2}= \sqrt{26}-\sqrt{2} $

Similarly, maximum value will be $\sqrt{26} +\sqrt 2$

(2) Using triangular inequalities, $|z-2i|=|(z-1+3i)-(5i-1)|\ge| |3-i(z-1)|-\sqrt{26}|$, which gives the same result.

Here's the solution in my text book, which is possibly incorrect but I'm not sure if wrong or the book.

$$ \begin{aligned} |3-i(z-1)| &=|i(z-1-3 i)|=|z-1-3 i| \\ &=|z-2 i+(-1-i)| \\ & \leq|z-2 i|+|1+i| \\ & \leq \sqrt{2}+\sqrt{2} \\ &=2 \sqrt{2} \end{aligned} $$ $\therefore$ Maximum value $=2 \sqrt{2}$

Answer 1

$|3-i(z-1)| = |z -(1 -3 i) |$

The $z$ in question can be written as $2i+ r e^{i \theta}$ with $r \in [0,\sqrt{2}]$, so the question is to maximise $|2i+ r e^{i \theta} -(1 -3 i) | = | 5i-1 + r e^{i \theta}|$.

Noting that $| 5i-1 + r e^{i \theta}| \le | 5i-1| + r$ (with equality when $\theta = \operatorname{arg} ( 5i-1 )$) then we get $\sqrt{26}+\sqrt{2}$.

Answer 2

Let $z=x+iy$ where $x,y$ are real so that $3-i(z-1)=3-i(x+iy-1)=3+y+i(1-x)$

Let $3+y+i(1-x)=r(\cos t+i\sin t)$

$|z-2i|^2=|x+iy-2i|^2=(r\sin t-1)^2+(r\cos t-5)^2=r^2+26-2r(\sin t +5\cos t)$

We need $$r^2+26-2r(\sin t +5\cos t)-2\le0$$

Now the roots of $$r^2+26-2r(\sin t +5\cos t)-2=0$$ are

$$\sqrt{26-(\cos t-5\sin t)^2}\pm\sqrt{2-(\cos t-5\sin t)^2}$$

$$\implies r\le\sqrt{26-0}+\sqrt{2-0}$$ the equality occurs if $\cos t-5\sin t=0$