Could anyone gimmie any clues with this please ?
By writing z + 2 + i = (z + 2i) + (2-i)
prove that
for z being in A where A= 1
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 !
A geometric method: |z-a| can be interpreted as the distance between points z and a in the complex plane.
Saying that z satisfies 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|= > 2 so i- 2 is outside that "washer". The shortest distance from i- 2 to the larger circle is while the longest distance, to the other side of the outer circle, is .