Chapter 6.8, Problem 11E

Interpretation Introduction

Interpretation:

For the complex vector field ${\dot{\text{z}}\text{ = z}}^{\text{k}}$, write the vector fields in both Cartesian and polar coordinates, and for the cases $\text{k=1,2,3,4}$, show that the origin is the only fixed point, and compute its index.

Concept Introduction:

The index of a closed curve C is an integer that measures the winding of the vector curve C is an integer that measures the winding of the vector field on C.

The index also provides information about a fixed point that might happen to lie inside the curve.

Index theory provides global information about the phase portrait

The index of the closed curve C can be defined as the net number of counterclockwise revolutions that are made by the vector field.

Answer to Problem 11E

Solution:

The vector field for ${\dot{\text{z}}\text{ = z}}^{\text{k}}$ in both Cartesian and polar coordinates, for the case $\text{k = 1}$, the field is $\dot{\text{z}}\text{ = x + iy}$ and ${\dot{\text{z}}\text{ = re}}^{\text{iθ}}$ respectively, and for $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$ in both Cartesian and polar coordinates, the field is $\dot{\text{z}}\text{ = x-iy}$ and ${\dot{\text{z}}\text{ = re}}^{\text{-iθ}}$ respectively.

The vector field for ${\dot{\text{z}}\text{ = z}}^{\text{k}}$ in both Cartesian and polar coordinates, for the case $\text{k = 2}$, the field is ${\dot{\text{z}}\text{ =x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{+2ixy}$ and ${\dot{\text{z}}\text{ =r}}^{\text{2}}{\text{e}}^{\text{2iθ}}$ respectively, and for $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$ in both Cartesian and polar coordinates, ${\dot{\text{z}}\text{ =x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{-2ixy}$ and ${\dot{\text{z}}\text{ =r}}^{\text{2}}{\text{e}}^{\text{-2iθ}}$ respectively.

The vector field for ${\dot{\text{z}}\text{ = z}}^{\text{k}}$ in both Cartesian and polar coordinates, for the case $\text{k = 3}$, the field is ${\dot{\text{z}}\text{=x}}^{\text{3}}{\text{-3xy}}^{\text{2}}\text{+i}\left({\text{3x}}^{\text{2}}{\text{y - y}}^{\text{3}}\right)$ and ${\dot{\text{z}}\text{=r}}^{\text{3}}{\text{e}}^{\text{3iθ}}$ respectively, and for $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$ in both Cartesian and polar coordinates, ${\dot{\text{z}}\text{=x}}^{\text{3}}{\text{-xy}}^{\text{2}}\text{-i}\left({\text{3x}}^{\text{2}}{\text{y-y}}^{\text{3}}\right)$ and ${\dot{\text{z}}\text{ =r}}^{\text{3}}{\text{e}}^{\text{-3iθ}}$ respectively.

It is to be shown that origin is the only fixed point.

The vector fields will have $\text{k}$ rotations on a loop enclosing the origin.

Explanation of Solution

Index theory provides global information about the phase portrait. The index of the closed curve C can be defined as the net number of counterclockwise revolutions that are made by the vector field. Mathematically, it can be expressed as:

${\text{I}}_{\text{c}}\text{=}\frac{\text{1}}{\text{2π}}{\left[\varphi \right]}_c$

Here, ${\text{I}}_{\text{c}}$ is the index of the closed curve, ${\left[\varphi \right]}_c$ is the net charge in $\varphi$ around C.

Consider complex vector fields as,

${\dot{\text{z}}\text{ = z}}^{\text{k}}$ And $\dot{\text{z}}\text{ = }{\left(\overline{\text{z}}\right)}^{\text{k}}$

Here, k is an integer and is greater than0. Also, $\overline{\text{z}}\text{ = x -iy}$ is the complex conjugate of z

Let, $\text{z= x + iy }$

The Euler’s formula is given as,

${\text{cosθ + i sin θ = e}}^{\text{iθ}}$

a)

Write the vector fields in both Cartesian and polar coordinates for different cases

Case 1)

$\text{k = 1}$

Solve for

Substitute, 1 for $\text{k}$ in ${\dot{\text{z}}\text{ = z}}^{\text{k}}$,

$\dot{\text{z}}\text{ = }{\left(\text{z}\right)}^1$

Substitute, $\text{z= x + iy }$ and solve,

$\dot{\text{z}}\text{ = x + iy}$

$\dot{\text{z}}\text{ = r cosθ + i }\left(\text{r sinθ}\right)$

$\dot{\text{z}}\text{ = r}\left(\text{cosθ + i sinθ}\right)$

Substitute, ${\text{cosθ + i sin θ = e}}^{\text{iθ}}$

${\dot{\text{z}}\text{ = re}}^{\text{iθ}}$

Here, $\text{r=}\sqrt{{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}}$

And, ${\text{θ = tan}}^{\text{-1}}\left(\frac{\text{y}}{\text{x}}\right)$

Thus, the Cartesian and polar co-ordinate for this vector field is $\dot{\text{z}}\text{ = x + iy}$ and ${\dot{\text{z}}\text{ = re}}^{\text{iθ}}$ respectively.

Solve for $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$

Substitute, 1 for $\text{k}$ in $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$

$\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{1}}$

Substitute $\overline{\text{z}}\text{ = x-iy}$

$\dot{\text{z}}\text{ = x-iy}$

This is the Cartesian co-ordinate for this vector field.

Solve for polar coordinates.

Substitute, $\text{r cos θ}$ for x, $\text{r sin θ}$ for $\text{y}$ in $\dot{\text{z}}\text{ = x-iy}$ and solve

$\dot{\text{z}}\text{ = r cos θ- i}\left(\text{r sin θ}\right)$

$\dot{\text{z}}\text{ = r}\left(\text{cos θ- isin θ}\right)$

Substitute, ${\text{e}}^{\text{-iθ}}\text{=}\left(\text{cos θ- isin θ}\right)$

${\dot{\text{z}}\text{ = re}}^{\text{-iθ}}$

Thus, the Cartesian and polar coordinates for this vector field are $\dot{\text{z}}\text{ = x-iy}$ and ${\dot{\text{z}}\text{ = re}}^{\text{-iθ}}$ respectively.

Case 2)

$\text{k=2}$

Solve for ${\dot{\text{z}}\text{ = z}}^{\text{k}}$

$\dot{\text{z}}\text{ = }{\left(\text{z}\right)}^2$

Substitute, $\text{z= x + iy }$ and solve,

$\dot{\text{z}}\text{ =}{\left(\text{x+iy}\right)}^{\text{2}}$

${\dot{\text{z}}\text{ =x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{+2ixy}$

This is the Cartesian co-ordinate for this vector field.

Solve for polar coordinates,

Substitute, $\text{rcosθ}$ for $\text{x}$ and $\text{rsinθ}$ for $\text{y}$

$\dot{\text{z}}\text{ =}{\left(\text{r cosθ+i}\left(\text{sinθ}\right)\right)}^{\text{2}}$

${\dot{\text{z}}\text{ =r}}^{\text{2}}{\left(\text{cosθ+isinθ}\right)}^{\text{2}}$

Substitute, ${\text{cosθ + i sin θ = e}}^{\text{iθ}}$

${\dot{\text{z}}\text{ =r}}^{\text{2}}{\text{e}}^{\text{2iθ}}$

Thus, the Cartesian and polar co-ordinate for this vector field are ${\dot{\text{z}}\text{ =x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{+2ixy}$ and ${\dot{\text{z}}\text{ =r}}^{\text{2}}{\text{e}}^{\text{2iθ}}$ respectively.

Solve for $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$

Substitute 2 for $\text{k}$ in $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$

$\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^2$

Substitute $\overline{\text{z}}\text{ = x-iy}$

$\dot{\text{z}}\text{ =}{\left(\text{x-iy}\right)}^{\text{2}}$

${\dot{\text{z}}\text{ =x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{-2ixy}$

This is the Cartesian co-ordinate for this vector field.

Solve for polar coordinates.

Substitute, $\text{r cos θ}$ for x, $\text{r sin θ}$ for $\text{y}$ in $\dot{\text{z}}\text{ = x-iy}$ and solve

$\dot{\text{z}}\text{ =}{\left(\text{rcosθ-i}\left(\text{rsinθ}\right)\right)}^{\text{2}}$

${\dot{\text{z}}\text{ =r}}^{\text{2}}{\left(\text{cosθ - isinθ}\right)}^2$

Substitute, ${\text{e}}^{\text{-iθ}}\text{=}\left(\text{cos θ- isin θ}\right)$

${\dot{\text{z}}\text{ =r}}^{\text{2}}{\text{e}}^{\text{-2iθ}}$

Thus, the Cartesian and polar coordinates for this vector field are ${\dot{\text{z}}\text{ =x}}^{\text{2}}{\text{-y}}^{\text{2}}\text{-2ixy}$ and ${\dot{\text{z}}\text{ =r}}^{\text{2}}{\text{e}}^{\text{-2iθ}}$ respectively.

Case 3)

$\text{k=3}$

Solve for ${\dot{\text{z}}\text{ = z}}^{\text{k}}$

$\dot{\text{z}}\text{ = }{\left(\text{z}\right)}^3$

Substitute, $\text{z= x + iy }$ and solve,

$\dot{\text{z}}\text{ =}{\left(\text{x+iy}\right)}^3$

${\dot{\text{z}}\text{=x}}^{\text{3}}{\text{-3xy}}^{\text{2}}\text{+i}\left({\text{3x}}^{\text{2}}{\text{y - y}}^{\text{3}}\right)$

This is the Cartesian co-ordinate for this vector field.

Solve for polar coordinates,

Substitute, $\text{rcosθ}$ for $\text{x}$ and $\text{rsinθ}$ for $\text{y}$

$\dot{\text{z}}\text{ =}{\left(\text{r cosθ+i}\left(\text{sinθ}\right)\right)}^3$

${\dot{\text{z}}\text{ =r}}^3{\left(\text{cosθ+isinθ}\right)}^3$

Substitute, ${\text{cosθ + i sin θ = e}}^{\text{iθ}}$

Thus, the Cartesian and polar co-ordinate for this vector field are ${\dot{\text{z}}\text{=x}}^{\text{3}}{\text{-3xy}}^{\text{2}}\text{+i}\left({\text{3x}}^{\text{2}}{\text{y - y}}^{\text{3}}\right)$ and ${\dot{\text{z}}\text{=r}}^{\text{3}}{\text{e}}^{\text{3iθ}}$ respectively.

Solve for $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$

Substitute, 3 for $\text{k}$ in $\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^{\text{k}}$

$\dot{\text{z}}\text{ =}{\left(\overline{\text{z}}\right)}^3$

Substitute $\overline{\text{z}}\text{ = x-iy}$

$\dot{\text{z}}\text{=}{\left(\text{x-iy}\right)}^{\text{3}}$

${\dot{\text{z}}\text{=x}}^{\text{3}}{\text{-xy}}^{\text{2}}\text{-i}\left({\text{3x}}^{\text{2}}{\text{y-y}}^{\text{3}}\right)$

This is the Cartesian co-ordinate for this vector field.

Solve for polar coordinates.

Substitute, $\text{r cos θ}$ for x, $\text{r sin θ}$ for $\text{y}$ in $\dot{\text{z}}\text{ = x-iy}$ and solve

$\dot{\text{z}}\text{ =}{\left(\text{rcosθ-i}\left(\text{rsinθ}\right)\right)}^{\text{3}}$

${\dot{\text{z}}\text{ =r}}^{\text{3}}{\left(\text{cosθ - i sinθ}\right)}^{\text{3}}$

Substitute, ${\text{e}}^{\text{-iθ}}\text{=}\left(\text{cos θ- isin θ}\right)$

${\dot{\text{z}}\text{ =r}}^{\text{3}}{\text{e}}^{\text{-3iθ}}$

Thus, the Cartesian and polar coordinates for this vector field are ${\dot{\text{z}}\text{=x}}^{\text{3}}{\text{-xy}}^{\text{2}}\text{-i}\left({\text{3x}}^{\text{2}}{\text{y-y}}^{\text{3}}\right)$ and ${\dot{\text{z}}\text{ =r}}^{\text{3}}{\text{e}}^{\text{-3iθ}}$ respectively.

b)

It is to be shown that origin is the only fixed point

Take case 1 as $\text{k=1}$

Substitute, $\text{0}$ for $\dot{\text{z}}$ in the vector field ${\dot{\text{z}}\text{ =re}}^{\text{iθ}}$

${\text{0=re}}^{\text{iθ}}$

$\text{r=0}$

Substitute, $\text{0}$ for $\dot{\text{z}}$ in the vector field ${\dot{\text{z}}\text{ =re}}^{\text{-iθ}}$

${\text{0=re}}^{\text{-iθ}}$

$\text{r=0}$

The index can be calculated as,

${\text{I}}_{\text{c}}\text{=}\frac{\text{1}}{\text{2π}}{\left[\varphi \right]}_{\text{c}}$

For the vector field ${\dot{\text{z}}\text{ =e}}^{\text{ kiθ}}$, substitute $\text{kθ}$ for $\varphi$ and take the limits from $\text{0 to 2}\pi$

$\begin{array}{l}{\text{I}}_{\text{c}}\text{=}\frac{\text{1}}{\text{2π}}{\left[\text{kθ}\right]}_{\text{0}}^{\text{2π}}\\ {\text{I}}_{\text{c}}\text{=}\frac{\text{1}}{\text{2π}}\text{k}\left(\text{2π-0}\right)\\ {\text{I}}_{\text{c}}\text{=k}\end{array}$

For the vector field ${\dot{\text{z}}\text{ =e}}^{\text{- kiθ}}$, substitute $\text{kθ}$ for $\varphi$ and take the limits from $\text{0 to 2}\pi$

$\begin{array}{l}{\text{I}}_{\text{c}}\text{=}\frac{\text{1}}{\text{2π}}{\left[\text{-kθ}\right]}_{\text{0}}^{\text{2π}}\\ {\text{I}}_{\text{c}}\text{=}\frac{\text{1}}{\text{2π}}\left(\text{-k}\right)\left(\text{2π-0}\right)\\ {\text{I}}_{\text{c}}\text{= -k}\end{array}$

c)

Let us generalize that this result for $\text{k>0}$, the index is calculated in the previous section comes out to be $\text{±k}$ in the vector fields ${\text{z}}^{\text{k}}$ and ${\left(\overline{\text{z}}\right)}^{\text{k}}$ respectively. This implies that the vector fields will have $\text{k}$ rotations on a loop enclosing the origin. Here, $\text{θ}$ goes through a rotation of $\text{2}\pi$ due to the k in the term ${\text{e}}^{\text{±kiθ}}$. Here, $\text{kiθ}$ is causing counterclockwise rotation and $\text{- kiθ}$ is causing clockwise rotation.