Complex analysis topic help Need explanation

[nate9228]

Homework Statement

This isn't a specific problem, but more of a type of problem I do not get. I am taking undergrad complex analysis, using the book by Bak and Newman. Its only a couple week in and I am having to spend a lot of time on it, last week I spent about 7 hours on the homework (which was only 10 problems. Anyways one of the types of problems I don't understand how to do is absolute value problems of the sort that goes "Identify/Describe the set of points that satisfy...". Examples would be "identify etc, \left|z\right|=Rex+1. Or \left|z-1\right|+\left|z+1\right|=4. Or even problems that state the same beginning but do not have absolute value in them such as 1/z= z conjugate. Is there an overarching way to look at these problems? I just had a hard time picturing entire sets of points in my head I guess; I am use to finding a single value.

Homework Equations

The Attempt at a Solution

 

[jbunniii]

One tactic is to try to express the set of points in terms of the real and imaginary parts of z, or in terms of the magnitude and phase angle of z. It's not always clear ahead of time which approach will work better, so this may take some experimenting.

Some general things to keep in mind: $|z|^2 = z \bar{z}$, $z + \bar{z} = 2\textrm{Re}(z)$, $z - \bar{z} = 2i\textrm{Im}(z)$.

So, for your first example, which I guess should read $|z| = \textrm{Re}(z) + 1$, if we write $z = x + iy$, then the equation becomes $|x + iy| = x + 1$. We can square this to obtain $|x + iy|^2 = (x + 1)^2$, or $x^2 + y^2 = x^2 + 2x + 1$. Simplifying, we get $y^2 = 2x + 1$, or $x = \frac{1}{2}(y^2 - 1)$, which is a parabola opening to the right with vertex at $x=-1/2, y=0$.

 

[jbunniii]

For equations like $|z-1| + |z + 1| = 4$, it's usually helpful to rearrange so that one absolute value is on each side: $|z - 1| = 4 - |z + 1|$. Then square both sides to get $|z-1|^2 = 16 - 8|z+1| + |z + 1|^2$. Then put $z = x + iy$ and simplify:
$$\begin{align}
|x - 1 + iy|^2 &= 16 - 8|x + 1 + iy| + |x + 1 + iy|^2 \\
(x-1)^2 + y^2 &= 16 - 8|x + 1 + iy| + (x+1)^2 + y^2 \\
8|x + 1 + iy| &= 16 + (x+1)^2 - (x - 1)^2 \\
8|x + 1 + iy| &= 16 + 4x \\
\end{align}$$
Now square both sides again and continue simplifying.

 

[Ray Vickson]

For equations like $|z-1| + |z + 1| = 4$, it's usually helpful to rearrange so that one absolute value is on each side: $|z - 1| = 4 - |z + 1|$. Then square both sides to get $|z-1|^2 = 16 - 8|z+1| + |z + 1|^2$. Then put $z = x + iy$ and simplify:
$$\begin{align}
|x - 1 + iy|^2 &= 16 - 8|x + 1 + iy| + |x + 1 + iy|^2 \\
(x-1)^2 + y^2 &= 16 - 8|x + 1 + iy| + (x+1)^2 + y^2 \\
8|x + 1 + iy| &= 16 + (x+1)^2 - (x - 1)^2 \\
8|x + 1 + iy| &= 16 + 4x \\
\end{align}$$
Now square both sides again and continue simplifying.

For $|z-1| + |z + 1| = 4$ it is easiest to look at the geometry. The term $|z-1|$ is the distance from z to p = -1 + 0i, while $|z-1|$ is the distance from z to q = +1 + 0i. Therefore, the left-hand-side is the sum of the distances to two fixed points, and that sum must be the constant 4. Can you recognize what type of curve that would be?

 

[nate9228]

Thanks guys that definitely helps. I'm sure I'll be back with other questions soon!