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

complex numberscomplex-analysisderivativesexponential function

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

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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 $e^{\phi + i\phi} = e^{(i+1)\phi}$.

Best Answer

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.

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