91Ó°ÊÓ

First, let's express \(z\) and \(\bar{z}\) in polar coordinates. We know that \(z=x+iy=re^{i\theta}\) and \(\bar{z}=x-iy=re^{-i\theta}\), where \(r=\sqrt{x^2+y^2}\) and \(\theta=\tan^{-1}\frac{y}{x}\). For \(k=1\) and \(\dot{z}=z^k\), we have: In Cartesian coordinates: \(\dot{z}=x+iy\) In polar coordinates: \(\dot{z}=re^{i\theta}\) For \(k=2\) and \(\dot{z}=z^k\), we have: In Cartesian coordinates: \(\dot{z}=(x+iy)^2 = (x^2-y^2) + i(2xy)\) In polar coordinates: \(\dot{z}=(re^{i\theta})^2 = r^2e^{i(2\theta)}\) For \(k=3\) and \(\dot{z}=z^k\), we have: In Cartesian coordinates: \(\dot{z}=(x+iy)^3 = (x^3-3xy^2) + i(3x^2y-y^3)\) In polar coordinates: \(\dot{z}=(re^{i\theta})^3 = r^3e^{i(3\theta)}\) Now, let's do the same for \(\dot{z}=(\bar{z})^k\): For \(k=1\) and \(\dot{z}=(\bar{z})^k\), we have: In Cartesian coordinates: \(\dot{z}=x-iy\) In polar coordinates: \(\dot{z}=re^{-i\theta}\) For \(k=2\) and \(\dot{z}=(\bar{z})^k\), we have: In Cartesian coordinates: \(\dot{z}=(x-iy)^2 = (x^2-y^2) - i(2xy)\) In polar coordinates: \(\dot{z}=(re^{-i\theta})^2 = r^2e^{-i(2\theta)}\) For \(k=3\) and \(\dot{z}=(\bar{z})^k\), we have: In Cartesian coordinates: \(\dot{z}=(x-iy)^3 = (x^3-3xy^2) - i(3x^2y-y^3)\) In polar coordinates: \(\dot{z}=(re^{-i\theta})^3 = r^3e^{-i(3\theta)}\)

A fixed point occurs when \(\dot{z}=z\) for \(z^k\) and \(\dot{z}=\bar{z}\) for \((\bar{z})^k\). Let's first examine \(\dot{z}=z^k\). For \(k=1\): \(\dot{z}=z\), so \(x+iy = x^1+iy^1\), which is true only at the origin (0,0). For \(k=2\): \(\dot{z}=z^2\), so \(x+iy = (x^2-y^2) + i(2xy)\), which implies \(x(x^2-y^2) + y(2xy) = 0\). Factoring out x and y gives us \(x(x^2+y^2) = 0\) and \(y(x^2+y^2) = 0\). This is true only at the origin (0,0). For \(k=3\): \(\dot{z}=z^3\), so \(x+iy = (x^3-3xy^2) + i(3x^2y-y^3)\), which implies \(x(x^3-3xy^2) + y(3x^2y-y^3) = 0\). Factoring out x and y gives us \(x(x^2+y^2)^2 = 0\) and \(y(x^2+y^2)^2 = 0\). This is true only at the origin (0,0). Similarly, for \(\dot{z}=(\bar{z})^k\): For all cases \(k=1,2,3\), the only fixed point is the origin (0,0). To calculate the index of the fixed point, we need to compute the Jacobian matrix and then compute its determinant. The Jacobian matrix for the complex vector field \(\dot{z}=z^k\) is: \[ J=\begin{bmatrix} \frac{\partial (u)}{\partial x} & \frac{\partial (u)}{\partial y}\\ \frac{\partial (v)}{\partial x} & \frac{\partial (v)}{\partial y} \end{bmatrix} \] For \(k=1\), the index is 1, for \(k=2\), the index is 2, and for \(k=3\), the index is 3. The same applies for \(\dot{z}=(\bar{z})^k\).

We can generalize these results for any positive integer \(k\). For the complex vector fields \(\dot{z}=z^k\) and \(\dot{z}=(\bar{z})^k\), the only fixed point is the origin (0,0), and the index of the fixed point is equal to \(k\). This happens because as we increase the power of \(k\), the fixed point at the origin becomes more "attractive," and its influence on the system's dynamics increases.