We begin by assuming that $\cos\theta+i\sin\theta$ can be expressed in the form $R\cos(\theta-\alpha)$: $$\begin{align} \cos\theta+i\sin\theta =&R\cos(\theta-\alpha)\\ =&R\cos\theta\cos\alpha+R\sin\theta\sin\alpha\\ \end{align}$$ $$\implies R\cos\alpha=1,~~R\sin\alpha=i$$ $$\implies R=\sqrt{1+(i)^2}=\sqrt{1-1}=0$$ Also, $$\tan\alpha=i\implies \alpha=\arctan{i}$$ So we finish with $$0\times\cos(\theta-\arctan{i})=\cos\theta+i\sin\theta$$ How can this be?
Thanks for your help.
If $\cos\theta+i\sin\theta=R\cos(\theta-\alpha)$, you cannot assume that $R$ and $\alpha$ are real, unless $\sin\theta=0$. What you proved is that there do not exist $R$ and $\alpha$ such that $R\cos\alpha=1$ and $R\sin\alpha=i$, because this implies $R=0$.
Let's see whether we can find them so that $\cos\theta+i\sin\theta=R\cos(\theta-\alpha)$ holds for every $\theta$.
With $\theta=0$, we get $1=R\cos(-\alpha)$; with $\theta=\pi/2$, we get $i=R\sin\alpha$. OK, they don't exist, because this implies $R^2\cos^2\alpha+R^2\sin^2\alpha=1+i^2=0$, but then $R^2=0$. Indeed $\cos^2\alpha+\sin^2\alpha=1$ for every $\alpha\in\mathbb{C}$.