Let $$y=\cos \phi+i\sin \phi \tag{1}$$

Differentiating both sides of equation (1) with respect to $\phi$, we get,

$$\begin{align} \frac{dy}{d\phi} &=-\sin \phi+i\cos \phi \\ &=i(\cos \phi-\frac{1}{i}\sin \phi) \\ &=i(\cos\phi+i\sin \phi) \\ &=iy \\ \implies\frac{1}{y}dy &=i\;d\phi \tag{2} \end{align}$$

Integrating both sides of equation (2), we get,

$$\begin{align} \int\frac{1}{y}dy&=\int i\;d\phi \\ \implies \ln(y)&=i\phi+c \tag{3} \end{align}$$

Substituting $\phi=0$ in equation (1), we get,

$$y=\cos 0+i\sin 0 \implies y=1 \tag{4}$$

Substituting $\phi=0$ and $y=1$ in equation (3) we get,

$$\ln(1)=c \implies c=0 \tag{5}$$

Substituting $c=0$ in equation (3) we get,

$$\begin{align} \ln(y)&=i\phi \tag{6}\\ \implies e^{i\phi}&=y \tag{7}\\ \therefore e^{i\phi}&=\cos \phi+i\sin \phi \tag{8} \end{align}$$

I found this proof in a book. I think that there is a problem. In $(3)$ and $(6)$, shouldn't $\ln(y)$ be $\ln|y|$ instead? Or is it that, for complex numbers, we do not take the absolute value of the number within the $\ln$?


2 Answers 2


Your proof is correct, but there are some hidden assumptions which your book has conveniently failed to mention.

In particular the key assumption is the definition of symbol $e^{iy} $ for $y\in\mathbb{R} $ or equivalently the definition of $\log z$ for $z\in\mathbb{C} $.

One approach is to define $e^z, z\in\mathbb{C} $ via the limit $$e^z=\lim_{n\to\infty} \left(1+\frac{z}{n}\right)^n\tag{1}$$ or via the series $$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots\tag{2}$$ Using any of the above definitions one can easily prove that $e^{iy} =\cos y+i\sin y$. In particular the series definition can be used with the proof you have provided. Using series definition one can prove that if $f(y) =e^{iy} $ then $f'(y) =if(y) $ and then we can use the function $$g(y) =f(y) (\cos y-i\sin y) $$ to establish $g'(y) =0$ so that $g$ is constant. Thus $g(y) =g(0)=1$ and $f(y) =\cos y+i\sin y$.

The proof using limit definition $(1)$ is more interesting and you may have a look at it. Both the approaches avoid integrals altogether.

  • $\begingroup$ I am sorry but i could not understand your point properly. Could please explain your point? $\endgroup$
    – MrAP
    May 1, 2018 at 7:16
  • $\begingroup$ @MrAP: your proof till $dy/d\phi=iy$ is fine. After that it is difficult to assimilate. There are many complications: $y$ in your proof is a complex variable and handling the integral $\int (1/y)dy$ requires some sort of complex analysis. This requires the definition of complex logarithm and its derivative. The problem is handled easily using the approaches I mentioned. $\endgroup$
    – Paramanand Singh
    May 1, 2018 at 15:41

Note that indefinite integrals defined algebraically deal with complex quantities. However, many elementary calculus textbooks write formulas such as

$$\int \frac{dx}{x} = \ln{|x|} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\: (4)$$

(where the notation $x$ is used to indicate that $x$ is assumed to be a real number) instead of the complex variable version

$$\int \frac{dz}{z} = \ln{z} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: (5)$$

where z is generically a complex number (but also holds for real $z$).


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