1. ## Complex Inequality

Could anyone gimmie any clues with this please ?

By writing z + 2 + i = (z + 2i) + (2-i)

prove that $\sqrt {5} -2 \le$ $|z + 2 + i|$ $\le \sqrt {5} +2$

for z being in A where A= 1 $\le |z+2i| \le 2$

2. Originally Posted by frodozoid
Could anyone gimmie any clues with this please ?

By writing z + 2 + i = (z + 2i) + (2-i)

prove that $\sqrt {5} -2 \le$ $|z + 2 + i|$ $\le \sqrt {5} +2$

for z being in A where A= 1 $\le |z+2i| \le 2$
By the triangle inequality, $\left|(z+2i)+(2-i)\right|\leqslant\left|z+2i\right|+\left|2-i\right|$

This then leads to the fact that the upper bound becomes $\left|z+2+i\right|=\left|(z+2i)+(2-i)\right|\leqslant\left|z+2i\right|+\left|2-i\right|=2+\sqrt{(-1)^2+2^2}=\color{red}\boxed{2+\sqrt{5}}$

Now, for the lower bound, $\left|\left|z+2i\right|-\left|2-i\right|\right|\leqslant\left|(z+2i)+(2-i)\right|=\left|z+2+i\right|$. This implies that $\left|\left|2-i\right|-\left|z-2i\right|\right|=\left|\sqrt{5}-2\right|={\color{red}\boxed{\sqrt{5}-2}}\leqslant\left|(z+2i)+(2-i)\right|=\left|z+2+i\right|$

Thus, ${\color{red}\boxed{\sqrt{5}-2}}\leqslant\left|z+2+i\right|\leqslant{\color{red }\boxed{\sqrt{5}+2}}$

3. Ah thanks Chris, I see what I had missed now.
I was forgetting to take the magnitude of 2-i in the correct way, this is what was confusing me. That's where the root 5 comes from yeah.
I had the first line written out but a stupid error like that and I was confused haha. Nice one man cheers !

4. A geometric method: |z-a| can be interpreted as the distance between points z and a in the complex plane.

Saying that z satisfies $1\le |(z+ 2i)- 0|\le 2$ means that z+ 2i is in the "washer" with center 0, inner radius 1, outer radius 2.

By "writing z+ 2+ i as z+ 2i+ 2-i" we can interpret |z+ 2+ i|= |(z+2i)- (i-2)| as the distance from the point z+2i to the point i-2. |i- 2|= $\sqrt{5}$> 2 so i- 2 is outside that "washer". The shortest distance from i- 2 to the larger circle is $\sqrt{5}- 2$ while the longest distance, to the other side of the outer circle, is $\sqrt{5}+ 2$.