Maximum and minimum of $|z_1-z_2|$ whereas $|z_2|=2,z_1=3+4i$

Question

What are the greatest and least values of $|z_1-z_2|$ whereas $|z_2|=2,z_1=3+4i$ & $z_1,z_2$ are complex numbers?

$|z_2|=2$

$z_1=3+4i$

$|z_1-z_2|=?$

Let $z_2=a+bi$

After doing some manipulation I got $|z_1-z_2|=\sqrt{29-(6a+8b)}$

Now I have to find out the minimum of $6a+8b$ to get the maximum of $|z_1-z_2|$ & vice versa.

But How can I find those min & max , I'm not getting any ideas!!

Answer 1

You may just use both the triangle and the reverse triangle inequality:

• $||z_2|- |z_1|| \leq |z_2 - z_1| \leq |z_2| + |z_1|$

So, you get

$$5-2 = 3 \leq |z_2 - z_1| \leq 5+2 = 7$$

Equality is reached if $z_1$ and $z_2$ are parallel.

Answer 2

If you draw both loci it will become clear that the maximum and minimum values are given by $$|3+4i|\pm2=5\pm2=3,7$$

Answer 3

WLOG $z_1=2(\cos t+i\sin t)$ where $t$ is real

$|z_1-z_2|^2=4+3^2+4^2-6\cos t-8\sin t$

Now $-\sqrt{a^2+b^2}\le a\cos t+b\sin t\le\sqrt{a^2+b^2}$

Answer 4

You have $|z_1|=2 \Rightarrow a^2+b^2=4$ and $|z_1-z_2|=\sqrt{(a-3)^2+(b-4)^2}$.

Note that $a^2+b^2=4$ and $(a-3)^2+(b-4)^2=r^2$ are the two circles, for which $r$ must be minimized and maximized for the point $(a,b)$. The points will be the tangent points (internally and externally) of the two circles. They lie on the line passing through their centers $(0,0)$ and $(3,4)$: $b=\frac{4a}{3}$. Plugging this into the first equation: $$a^2+\left(\frac{4a}{3}\right)^2=4 \Rightarrow a=\pm \frac{6}{5} \Rightarrow b=\pm \frac{8}{5}.$$ You can find the min/max now.