# Graphing a function under a complex mapping

1. Jan 26, 2017

### cragar

1. The problem statement, all variables and given/known data
Illustrate the mapping of $f(z)=z+\frac{1}{z}$
for a parametric line.
3. The attempt at a solution
the equation for a parametric line is $z(t)=z_0(1-t)+z_1(t)$
so I plug z(t) in for z in f(z), but I dont get an obvious expression on how to graph it,
I tried manipulating it,
Was also wondering if I should represent $f(z)=\frac{(z-i)(z+i)}{z}$

Last edited by a moderator: Jan 26, 2017
2. Jan 26, 2017

### Ray Vickson

Pick some numerical values $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, then look at $f(z) = z + 1/z$ at $z = z_1 (1-t) + z_2\, t$:
$$\begin{array}{rcl}f(z) &=&z_1 (1-t) + z_2\, t + \frac{1}{z_1 (1-t) + z_2\, t}\\ &=& \displaystyle (x_1+ i y_1)(1-t) + (x_2 + iy_2) t + \frac{1}{ (x_1+ i y_1)(1-t) + (x_2 + iy_2) t } \end{array}$$
After some algebra this will have the form $A(t) + i B(t)$ for some functions $A, B$, so you get a parametric curve of the form $x = A(t)$, $y = B(t)$ to plot.

Last edited: Jan 27, 2017
3. Jan 27, 2017

### cragar

ok thanks, I also need to do it for a circle, For a circle the
equtation is $z(t)=re^{it}$
so If I plug this into f(z) I get , and im assuming r=1 for this $e^{it}+e^{-it}$
which is 2cos(t), so then Ijust graph 2*cos(t) as my answer.

4. Jan 27, 2017

### haruspex

That does not follow the recipe Ray gave you. He explained that you should plot y against x, not f against t.
Also, it would be better to avoid assuming r=1.