# Squaring a complex exponential that represents a real number

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.

• Something, something, branch cuts. Exponentiation is a multivalued thing in $\mathbb{C}$. Apr 29, 2021 at 16:16
• @SeanRoberson No, exponentiation is not multivalued. Apr 29, 2021 at 16:19
• @JoséCarlosSantos ah, I think Sean's probably thinking of logarithms then Apr 29, 2021 at 16:21
• @jumbot Yes, perhaps. Apr 29, 2021 at 16:21
• "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?" Could you explain and give an example where it "breaks down". So far as I can tell you example shows it works just fine. Apr 29, 2021 at 16:23

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 $$1^2\ne\operatorname{Re}\bigl((1+i)^2\bigr)=0$$.

• Ok, thanks. But what is the typical way to deal with exponents when using complex numbers to represent real ones? Apr 29, 2021 at 16:18
• There is no general rule, since the map $z\mapsto\operatorname{Re}(z)$ does not behave well with respect to multiplication and exponentiation. Apr 29, 2021 at 16:21
• Cool, so just convert it back to real numbers, do the exponentiation, and then put it back into the complex plane if necessary? Sounds like a plan. Thanks for your help! Apr 29, 2021 at 16:22
• @QuantumMechanic In those domains, you're actually considering complex sequences or functions as a vector space over $\Bbb R$, so addition and (real) scalar multiplication are the allowed operations. In this framework, $\Bbb C$ is essentially just $\Bbb R^2$, and the Euler formula is used to "convert" between these vectors and true complex values. Apr 30, 2021 at 2:26
• @AlexJones I agree, just mentioning that being unaware of the pitfalls of this mapping leads to similar mistakes when trying to take products of the functions in such contexts. Apr 30, 2021 at 12:40

The premise of the title is incorrect, and that may be the source of the confusion. When we write

$$\exp(it) = \cos t + i\sin t \in \Bbb C$$

this is not a complex number representing a real number. It is a complex number representing two real numbers.

A complex number representing one real number would be written as

$$\exp(ik\pi) = a + i\cdot 0 \in \Bbb C$$

and these numbers multiply and square as you would expect.

On the other hand, if you want compound numbers that represent two real numbers, like

$$(a,b)\in \Bbb R^2$$

then the only way to have multiplication (and exponentiation) preserve both values is to have a componentwise multiplication like the Hadamard product. On the other hand, $$\Bbb C$$ is differentiated from other structures on $$\Bbb R^2$$ precisely by its multiplication, which must have

$$(0,1)\cdot (0,1) = (-1,0) \neq (0,1)$$

and is therefore not componentwise.

For this particular example, the fact that

$$\exp((0,1)\cdot t) = (\cos t,\sin t)$$

holds at all is actually a result of the definition of complex multiplication. For $$\Bbb R^2$$ with a Hadamard product, we have

$$\exp((0,1)\cdot t) = (0,\exp t)$$

while for dual numbers (which have $$(0,1)\cdot (0,1) = 0$$), we have

$$\exp((0,1)\cdot t) = (1,t)$$

Interestingly, dual numbers actually do preserve the first component ("real part") through multiplication (and exponentiation), because

$$(a,x)\cdot (b,y) = (ab,ay+bx)$$

so it is possible to preserve one coordinate without componentwise multiplication, but the product on $$\Bbb C$$ doesn't do this.