Question

How to determine and sketch the complex plane |z+i| >= |z-i|?

Answer 1

doesnt this just mean \(b>0\) ?

Answer 2

or rather \(b\ge 0\)

Answer 3

Sorry I'm quite new to complex analysis, what is b again?

Answer 4

if you have \(z=a+bi\) then
\(|a+bi+i|=|a+(b+1)|=a^2+(b+1)^2=\sqrt{a^2+b^2+2b+1}\)

Answer 5

similarly \(|z-i|=\sqrt{a^2+b^2-2b+1}\)

Answer 6

Oh I see what you did there, but what does it mean by determine the plane?

Answer 7

solving the inequality gives \(b\geq 0\)

Answer 8

where in the plane is \(im(z)=b\geq 0\) ?

Answer 9

so would the graph by jw vs \[\sigma \]?

Answer 10

upper half plane

Answer 11

whoa okay. So firstly you change z into a+bi, then solve for b then find where b is located?

Answer 12

Is the graph still jw vs sigma

Answer 13

bc it says imaginary axis sorry for the questions

Answer 14

i have no idea what "jw vs \(\sigma\) "means

Answer 15

oh I see, what are the axis of the graph?

Answer 16

Oh gotcha thank you!