Another one:
How would you graph this:
View attachment 41060
You can use
![](https://latex.codecogs.com/png.latex?\bg_white z=x+iy)
to show it rearranges to
![](https://latex.codecogs.com/png.latex?\bg_white |y| < 1)
, and so is the region
![](https://latex.codecogs.com/png.latex?\bg_white -1 < \Im{(z)} < 1)
.
However, thinking of the expression as a vector, the interpretation of the statement is:
![](https://latex.codecogs.com/png.latex?\bg_white z)
is positioned such that the distance from
![](https://latex.codecogs.com/png.latex?\bg_white \bar{z})
to
![](https://latex.codecogs.com/png.latex?\bg_white z)
is less than 2 units
Since
![](https://latex.codecogs.com/png.latex?\bg_white \bar{z})
is located at the reflection of
![](https://latex.codecogs.com/png.latex?\bg_white z)
in the real axis,
![](https://latex.codecogs.com/png.latex?\bg_white z - \bar{z})
is a purely imaginary vector with its midpoint on the real axis.
So, if the length of this vector must be less than 2 units, then
![](https://latex.codecogs.com/png.latex?\bg_white z)
must lie within 1 unit (vertically) from the real axis.
Thus, the region
![](https://latex.codecogs.com/png.latex?\bg_white -1 < \Im{(z)} < 1)
covers all possible solutions for
![](https://latex.codecogs.com/png.latex?\bg_white z)
.
---
These kinds of vector interpretations can make more algebraically complicated problems easier.
For example: Find the maximum value of
![](https://latex.codecogs.com/png.latex?\bg_white |z|)
given that
![](https://latex.codecogs.com/png.latex?\bg_white |z - \bar{z}|<2)
and
![](https://latex.codecogs.com/png.latex?\bg_white |z + \bar{z}|<2)
.
The answer is
![](https://latex.codecogs.com/png.latex?\bg_white |z| < \sqrt{2})
, arising from the four values
![](https://latex.codecogs.com/png.latex?\bg_white z = 1\pm i)
and
![](https://latex.codecogs.com/png.latex?\bg_white z = -1\pm i)
that correspond to the four corners of the square that defines the region within which
![](https://latex.codecogs.com/png.latex?\bg_white z)
must lie.