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

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!!

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.

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

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

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.