3.53

For this problem, \(f(z)=z^2\)
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Show that the image under \(f\) of a circle centered at the origin is a circle centered at the origin.
Let \(z\) be any point on an arbitrary circle centered at the origin with radius \(r \neq 0\). Thus the equation of our point in polar coordinates is \(z=re^{i\theta}\). Considering our function, \(f(z)=z^2\), we have \(z^2=(re^{i\theta})^2\) ie \(r^2e^{2i\theta}\). Thus \(z^2\) lies on the circle of radius \(r^2\) centered at \(0\). Conversely, if \(w=r^2e^{i\phi}\) is a point on this second circle, then let \(z=...\). Then \(z^2=w\). Then by this function in polar coordinates would be \(r^2e^{2i\theta}+0\), where the +0 denotes that the center of the circle was not shifted by the function.

Show that the image under \(f\) of a ray starting at the origin is a ray starting at the origin.
Let \(R\) be an arbitrary ray centered at the origin with the equation \(re^{i\theta}\) with a fixed \(\theta\) where \(-\pi \leq \pi\) and \(0 \leq r \leq \infty\). Thus by our function \(f(z)=z^2\) gives us \(r^2e^{2i\theta}\). Thus \(R^2\) would have magnitude \(r^2\) and an argument of \(2\theta\). Since our function, \(f(z)\), does not shift the beginning point of \(R\) from the origin, \(f(R)\) still begins at the origin.

Let \(\textsf{T}\) be the figure formed by the horizontal segment from \(0\) to \(2\), the circular arc from \(2\) to \(2i\), and then the vertical segment from \(2i\) to \(0\). Draw \(\textsf{T}\) and \(f(\textsf{T})\).
See image below that is not rotatible.....

Is the right angel at the origin in part (c) preserved? Is something wrong here? No, the right angle is not preserved.
I do not believe something is wrong, since when you multiply two angles on the complex plane you add the arguments, so 90+90=180. No problem.