Discussing $\frac{d}{d\theta}e^{i\theta}$ aka cis before complex derivatives and complex exponential

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

Definition of $e^{i \theta}$ (or cis in other texts)

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About Prop 1.3f, how is it possible to discuss derivative of $e^{i \theta}$ before both defining derivatives of complex functions (Ch2) (including functions of a real variable I think!) and defining the complex exponential (Ch3)?

In particular, the proof of Prop 1.3f seems to assume linearity of the derivatives of complex functions.

There's even this exercise later on: Exer 1.6b

I know how to do this with Ch3's definition of the complex exponential. I don't believe this is possible to do with only Ch1 even if we write $e^{\phi + i\phi} = e^{(i+1)\phi}$.

Answer 1

I believe it's a mistake in the way that Mark Viola describes.

The correction to the mistake is I think as follows:

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I suspect a definition (*) such as

$$\frac{d}{d\theta}x(\theta)+iy(\theta) := \frac{d}{d\theta}x(\theta)+i\frac{d}{d\theta}y(\theta) \tag{1}$$

instead of something like 'we can treat $i$ as constant.

I think $(1)$ can be viewed as natural if you consider $z(\theta):=x(\theta)+iy(\theta) \in \mathbb C$ as a vector $\in \mathbb R^2$ s.t.

$$z(\theta) = [x(\theta),y(\theta)] \in \mathbb R^2$$

Then

$$z'(\theta)= [x'(\theta),y'(\theta)]$$

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(*) Definition is in Ch2.3. see the 2 red boxes below.

Answer 2

As the author states, this is just a definition of a function of $\phi$ and could have been denoted $q(\phi)$. This notation is chosen because it will accord with our definition of real and complex exponentiation. It can be hard to look at $e^{i\phi}$ and remember (until chapter 3) that you don't know this is the complex exponential, you just know it is this function of $\phi$.

The properties in $1.3$ can all be verified directly from the definition and the usual trig identities. In particular for $1.3f$ we have $$\frac d{d\phi}e^{i\phi}=\frac d{d\phi}(\cos \phi +i\sin \phi)=-\sin \phi+i\cos \phi=ie^{i\phi}$$ Note that the rule for differentiating an exponential was not used.

The challenge I see for $1.6b$ comes from mixing the real and imaginary numbers in the exponential. Even if you have already defined the exponential function for real arguments, we need to define $e^{a+bi}=e^ae^{bi}$ and I don't see a definition of that unless you have defined the sine and cosine of imaginary numbers. If you are given that definition and have the derivative of the real exponential you can prove what is desired.

Answer 3

When we define $e^{i\phi} = \cos \phi +i\sin \phi$ the only information from complex numbers is that $i$ is a constant whose square is $-1$

You can use the algebra of the complex field without using differentiation theorems of complex variables.

When we prove $\frac {d}{dz}z^2=2z$ using the definition of derivative we are using complex variables, but when we use $\frac {d}{d\phi }(cos \phi + i\sin \phi)= -\sin \phi + i \cos \phi $ we are just using differentiation rules of real vlued functions, $\sin x$ and $\cos x$ with the algebra of complex numbers.