1. ## complex plane

say if i have (p) |z+2|=4 and (q) Re(z)=0 which are two different lines in the hyperbolic plane and i want to find the point of intersection p interset q how would i do this?

2. Hyperbolic plane! You mean complex plane. Note $|z+2|=4$ is the circle $(x+2)^2 + y^2 = 4$ while $\Re (z) = 0$ is the line $x=0$.

3. sorry im rubbish i meant hyperbolic lines, ok so if i take what you said would i just substitute x=0 to get the intersection. i get confused as to how you get the circle from |z+2|=4

z=x+iy
| x+iy + 2|=4 group real and imaginary parts then mod

sqrt( ((x+2)^2) + (iy)^2)=4
and so on..
to get
((x+2)^2) - y^2 =16

where am i going wrong

4. Originally Posted by skystar
sorry im rubbish i meant hyperbolic lines, ok so if i take what you said would i just substitute x=0 to get the intersection. i get confused as to how you get the circle from |z+2|=4

z=x+iy
| x+iy + 2|=4 group real and imaginary parts then mod

sqrt( ((x+2)^2) + (iy)^2)=4 Mr F says: And here's the mistake. ${\color{red}|a + ib| = \sqrt{a^2 + b^2}}$, NOT ${\color{red}\sqrt{a^2 + (ib)^2}}$.
and so on..
to get
((x+2)^2) - y^2 =16

where am i going wrong
|z + 2| = 4 <=> |z - (-2)| = 4. So the geometric interpretation is the set of all points in the complex plane that are a distance 4 from z = -2. This is a circle of radius 4 and centre at z = -2, that is, (-2, 0).