Let C be the closed disc of radius 2 centred at $3+3i$.

Question

Let C be the closed disc of radius 2 centred at $3+3i$. The line $y=x$ splits C into an upper-left semi-disc and a lower-right semi-disc. Find a clockwise parametrization of the boundary of the upper-left semi-disc so that the line segment diameter is followed as $0 \leqslant t \leqslant 0.5$, then the upper-left semi-circle is traced as $0.5 \leqslant t \leqslant 1$.

So I know that $z_0=3+3i$ and $r=2$, and the formula used to parametrize a semi-circle is $z(t)=z_0+re^{it}$. Since we're dealing with clockwise orientation, $\pi$ is negative, so we get $z(t)=(3+3i)+2e^{-\pi it}$

For a line segment, I know the formula for parametrization is $z(t)=z_1+t(z_2-z_1)$, where $z_1$ is the start point and $z_2$ is the terminal point, but I'm having trouble figuring out how to parametrize the line segment and was wondering how to go about doing it.

Answer 1

First of all, the circle is described by $(x-3)^2+(y-3)^2=4$ so when this intersects $y=x$, this implies $(x-3)^2=2$. Therefore, $x=y=\sqrt{2}+3$ and $x=y=-\sqrt{2}+3$ are the two solutions.

To your point about the parameterization of a line, the initial point is $z_0=(\sqrt{2}+3) + (\sqrt{2}+3)i$, while the end point is $z_1=(-\sqrt{2}+3)+(-\sqrt{2}+3)i$. The parameterization is then $\gamma_1(t)=z_0+2t(z_1-z_0).$

For the next part, consider the upper left circle. You want, at $t=.5$, to be at $3+3i+2e^{5\pi/4i}$, and at $t=1$, to be at $3+3i+2e^{\pi/4i}$. This is can be found by considering the line connecting $\theta=5\pi/4$ to $\theta=\pi/4$. If it was $t=0$ to $t=.5$, the technique could be similar to the one above. All you have to do is shift $t$ to the right by $.5$. Then throw this line in the exponential to form the second curve $\gamma_2(t)$.