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.

  • $\begingroup$ Something, something, branch cuts. Exponentiation is a multivalued thing in $\mathbb{C}$. $\endgroup$ Apr 29, 2021 at 16:16
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    $\begingroup$ @SeanRoberson No, exponentiation is not multivalued. $\endgroup$ Apr 29, 2021 at 16:19
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    $\begingroup$ @JoséCarlosSantos ah, I think Sean's probably thinking of logarithms then $\endgroup$ Apr 29, 2021 at 16:21
  • $\begingroup$ @jumbot Yes, perhaps. $\endgroup$ Apr 29, 2021 at 16:21
  • $\begingroup$ "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. $\endgroup$
    – fleablood
    Apr 29, 2021 at 16:23

2 Answers 2


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

  • $\begingroup$ Ok, thanks. But what is the typical way to deal with exponents when using complex numbers to represent real ones? $\endgroup$ Apr 29, 2021 at 16:18
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    $\begingroup$ There is no general rule, since the map $z\mapsto\operatorname{Re}(z)$ does not behave well with respect to multiplication and exponentiation. $\endgroup$ Apr 29, 2021 at 16:21
  • $\begingroup$ 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! $\endgroup$ Apr 29, 2021 at 16:22
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    $\begingroup$ @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. $\endgroup$
    – Alex Jones
    Apr 30, 2021 at 2:26
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    $\begingroup$ @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. $\endgroup$ 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.


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