Why can't e^iπ be written in the form Rcos(x-a)

So, I understand that $a\sin{x} + b\cos{x}$ can usually be written in the form $R\cos{(x-a)}$, where $R = \sqrt{a^2 + b^2}$ and $a$ is equal to blah blah blah..
But since $e^{i\pi} = \cos\theta +i \sin\theta$, $R = \sqrt{1 + i^2} = 0$

Obviously algebraically, I understand why it's $0$, but intuitively, why can't it be written in this form? I've tried it for other coefficients of $i$ and it seems that $\cos(a)$ must be imaginary [is that possible?]. I just found it interesting and would kind of like an explanation as to why it only works for real numbers, thanks

e^(ipi) = cos pi + isin(pi) = cos pi = cos (pi - 0) so it's already in that form with R = 1 and a = 0

I actually thought you could probably do this if you allowed R and a to be complex, but turns out it doesn't work:

We want to find R, a such that

e^iz = cos z + i sin z = R (cos z cos a + sin z sin a)

setting z = 0 we have R cos a = 1
setting z = pi/2 we have R sin a = i
dividing we find tan a = i, that is, sin a /cos a = i and so i sin a / cos a = -1.

Since e^ia - e^-ia = 2i sin a and e^ia+e-ia = 2 cos a, we must have $\dfrac{e^{ia} - e^{-ia}}{e^{ia} + e^{-ia}} = -1$. Writing s = ia, this is:

$\dfrac{e^s - e^{-s}}{e^s+e^{-s}} = -1$ from which we see we must have $-s = \infty$ and so $a=i\infty$. Which isn't terribly helpful (I'm not sure if you can make it work if you're prepared to deal with this quantity, but I can't see it being fruitful).

Not sure if this sheds any light on anything to be honest...

What are your $a$ and $b$ here?

So the whole attempt fails for *any* complex number, no?

It's a very nice question though.

If you go back to the start, we are trying to solve for $R$ and $\alpha$ in

$a \cos(z) + b \sin(z) = R \cos(z - \alpha)$

However, I think that you have changed the original question posed the OP, who is an A level student, I think. As originally posed, I think that it is correct to state that there are no solutions, as he was explicitly looking for a solution based on $re^{ix} = r(\cos x +i \sin x)$ i.e. with $a=1,b=i$

$a = R \cos(\alpha)$
$b = R \sin(\alpha)$
$\alpha = \arctan(b/a)$ and $R^2 = a^2 + b^2$

As an extra credit exercise: do the values of $(a, b)$ that fail to give an answer for $\alpha$ (the "poles" of $\arctan(z)$) corresond to the values that fail to give a non-zero answer for $R$?