I was asked to determine the degree of the antipodal circle function $f(\theta)=\theta + \pi$. I believe the degree of the function is $1$, through pictures and illustration, but I will like to prove this using homotopy.
So given our degree $1$ map $C_{-1}(\theta):S^1 \rightarrow S^1$ is given by $$f(\cos(\theta)+i\sin(\theta))=\cos(\theta)+i\sin(\theta)$$.
Could I use as a homotopy $H(x,t)=(\cos(\theta)+i\sin(\theta))e^{t\pi i} $. If so how will prove it is continuous.
Yes, you can. However, you can simplify your formula by writing $$H(z,t) = z\cdot e^{\pi i t} .$$ Note that $S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. The map $H$ is continuous because complex multiplication is continuous and the exponential function is continuous.
If you want you can also write $S^1 = \{ (x,y) \in \mathbb R^2 \mid x^2 + y^2 = 1\}$ and $$H(x,y,t) = (x\cos(\pi t) - y \sin(\pi t), x \sin(\pi t) + y \cos(\pi t)) .$$