Graphing a function under a complex mapping

[cragar]

Homework Statement

Illustrate the mapping of $f(z)=z+\frac{1}{z}$
for a parametric line.

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 don't get an obvious expression on how to graph it,
I tried manipulating it,
Was also wondering if I should represent $f(z)=\frac{(z-i)(z+i)}{z}$

 

[Ray Vickson]

Homework Statement

Illustrate the mapping of $f(z)=z+\frac{1}{z}$
for a parametric line.

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 don't get an obvious expression on how to graph it,
I tried manipulating it,
Was also wondering if I should represent $f(z)=\frac{(z-i)(z+i)}{z}$

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.

 

[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 I am assuming r=1 for this $e^{it}+e^{-it}$
which is 2cos(t), so then Ijust graph 2*cos(t) as my answer.

 

[haruspex]

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 I am assuming r=1 for this $e^{it}+e^{-it}$
which is 2cos(t), so then Ijust graph 2*cos(t) as my answer.

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.