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.
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 $\gamma_2(t)$.