Find and classify the fixed points of each of the following functions:

- $\displaystyle f(z) = \frac{z^2}{2}+\frac{5 z}{2}+1$
- $\displaystyle f(z) = z^2+(2+i) z+i$
- $\displaystyle f(z) = \frac{1}{3}z + 6z^3$

An archived instance of discourse for discussion in undergraduate Complex Dynamics.

mark

Find and classify the fixed points of each of the following functions:

- $\displaystyle f(z) = \frac{z^2}{2}+\frac{5 z}{2}+1$
- $\displaystyle f(z) = z^2+(2+i) z+i$
- $\displaystyle f(z) = \frac{1}{3}z + 6z^3$

Rick

Let $f(z)=\frac{z^2}{2}+\frac{5z}{2}+1$.

Then to find the fixed points of $f$, $$z=\frac{z^2}{2}+\frac{5z}{2}+1 \rightarrow 0=\frac{z^2}{2}+\frac{3z}{2}+1 \rightarrow 0=z^2+3z+2=(z+2)(z+1).$$

Thus the fixed points of f are located at $z=-2$ and $z=-1.$

RedCrayon

@Rick We should also classify these points.

Note that $$f'(z)=z+5/2$$

So $f'(-2)=1/2$ and $f'(-1)=3/2$.

Thus, the point at $z=-2$ is an attractive fixed point and the point at $z=-1$ is a repulsive fixed point.

Yousername

Let $f(z)=1/3z+6z^3$,

Then, $0=-2/3z+6z^3$ --> $0=z(-2/3+6z^2)$

So, the fixed points of $f$ are $z=0$, $z=1/3$, and $z=(1/3)i$.

To classify these points we us the equation $f'(z)=1/3+18z^2$.

When $z=0$, $f'(z)=1/3$, so there is an attractive fixed point here.

When $z=1/3$, $f'(z)=1/3+18(1/9)=2 +1/3$, so there is a repulsive fixed point here.

When $z=(1/3)i$, $f'(z)=1/3+18(-1/9)=-1.6666$, so there is no fixed point here (or is it undefined? I'm not sure).

Am I doing this right?

**edit**

The third fixed point is -1/3, not (1/3)i. When $z=(1/3)$, $f'(z)=1/3+18(1/9)=2 +1/3$, so there is a repulsive fixed point here as well.

mark

@Yousername Close! I think you've got an error in your computation of the fixed point. I think the fixed points are $0$, $1/3$ and $-1/3$ - *not* $(1/3)i$. Your classification is correct for the fixed points that are correct.

Captain_Flapjack

So, for part $b$, where $f(z)=z^2+(2+i)z+i$, I should be looking for two fixed points as the roots of the $2nd$ order solution to $f(z)-z$:

$$

f(z)-z=z^2 +(1+i)z+i

$$

Using the good ol' quadratic formula I got the roots that at first looked not so nice:

$$

z=\frac{-1-i}{2}\pm\frac{\sqrt{-2i}}{2}

$$

Using Euler's identity I got the inside:

$$

\sqrt{-2i}=\left(2e^{\frac{-\pi}{2}}\right)^{\frac{1}{2}}=\sqrt{2}e^{\frac{-\pi}{4}}=1-i

$$

So my roots are $\frac{-1-i}{2}\pm\frac{1-i}{2}$ or $-1$ and $-i$, which I tested by evaluating them using $f(z)$. Sure enough, these are the fixed points of my function!

Also, if you compute the derivatives you find

$$f'(-1) = i \: \text{ and } \: f'(-i) = 2-i.$$

Thus, the points are neutral and repulsive.