Max and min of complex locus $|z-2-i|=1$

Question

I do not understand the process that is required in order to find the max and min values of $|z|$ in $|z-2-i|=1$. My textbook implies to use inspection which I find a bit confusing.

I tried to sketch out the curve but I have little idea on where the minimum and maximum values are.

Personally, I find A and B to be the intuitive places to check for a minimum, but they don't seem to be the right answers.

Additionally, I am conflicted between C or D (D which is somewhere between $(2,2)$ and C on the circle) is the maximum value. It seems like it could swing either way. Could you pelase explain intuitively how I should determine where the max and min are?

Answer 1

The minimal value of $\lvert z\rvert$ is reached at the point $P$ of the circle which is closest to the origin and the maximal value of $\lvert z\rvert$ is reached at the point $Q$ of the circle which is furthest from the origin. These points are$$P=2-\frac2{\sqrt5}+\left(1-\frac1{\sqrt5}\right)i\text{ and }Q=2+\frac2{\sqrt5}+\left(1+\frac1{\sqrt5}\right)i$$respectively.

Answer 2

Using the reverse triangle inequality you can derive maximum and minimum analytically:

$$|z-2-i|=1 \Rightarrow | |z| - |2+i| | \leq 1 \Rightarrow -1 \leq |z| - \sqrt{5} \leq 1 \Rightarrow \boxed{-1 +\sqrt{5} \leq |z| \leq 1 + \sqrt{5}}$$

Answer 3

Regarding those 4 points, you can do a direct computation. $$|A|= \sqrt{1^2+1^2} = \sqrt2\approx 1.4\\ |B|= 2\\|C|=\sqrt{1^2+3^2}=\sqrt{10}\approx 3.16 \\|D|=| 2+i + e^{i\pi/4}| = \sqrt{\left (2 + \frac{1}{\sqrt2}\right )^2+ \left (1 + \frac{1}{\sqrt2}\right )^2} \approx 3.2$$

Answer 4

$|z|$ is the distance from the origin to $z$. Geometrically, given a circle $S$ and a point $p$ outside the circle, the maximum and minimum distance from $p$ to $S$ are equal to the distance from $p$ to the center of the circle plus the radius and minus the radius, respectively.

The center of the circle is at $2+i$, which is $\sqrt5$ away from the origin, and the radius of the circle is $1$. The maximum and minimum distances from the origin to the circle are therefore $\sqrt5\pm 1$.

Answer 5

WLOG $z-2-i=\cos t+i\sin t$ where $t$ is real

$$z=2+\cos t+i(1+\sin t)$$

$$|z|^2=2^2+1^2+2(2\cos t+\sin t)=5+2\sqrt{2^2+1^2}\cos\left(t-\arccos\dfrac2{\sqrt5}\right)$$

Now $-1\le\cos\left(t-\arccos\dfrac2{\sqrt5}\right)\le1$