Often, complex exponential functions are used to represent trigonometric functions, since
$$
e^{i\theta} \equiv \cos\theta + i\sin\theta .
$$
Thus, if for example I want to express the quantity $\cos x$, I might write:
$$
\cos x \equiv \Re\left\{e^{i x}\right\} .
$$
I'm told that I can manipulate the LHS just like I would the RHS, and at the end just take the real part to get the same answer as other methods, but I have come across some trouble.
Let's say I wanted to square the LHS to get $\cos^2 x$. On the RHS, this would give me:
$$
\begin{align}
e^{2ix} &= (\cos x + i \sin x)^2 \\
&= (\cos^2x – \sin^2 x + 2i\cos x \sin x) \\
\implies \Re\{e^{2ix}\} &= \cos^2 x – \sin^2 x
\end{align}
$$
Now, of course I recognise that the RHS is the identity for $\cos 2x$, which makes complete sense since $e^{2ix} \equiv e^{i(2x)}$. My question then is, why do the rules suddenly break down as soon as I attempt to square my complex exponential as I would my trig function? And what are the most conventional steps to take to work around this? Many thanks.
Best Answer
The problem lies in the fact that you cannot deduce from $a=\operatorname{Re}(z)$ that $a^2=\operatorname{Re}(z^2)$, which is what you did. For instance, $1=\operatorname{Re}(1+i)$, but $1^2\ne\operatorname{Re}\bigl((1+i)^2\bigr)=0$.