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I am curious how mathematicians came to develop complex exponentiation. How is the rule for complex exponentiation derived?

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A motivation was, for instance, understanding what $\log(-1)$ should be. Granted that $e^r\ne-1$, for real $r$, Euler tried to understand what $e^{ir}$ should be. Then he did \begin{align} e^{ir}&= 1+\frac{ir}{1!}+\frac{(ir)^2}{2!}+\frac{(ir)^3}{3!}+\frac{(ir)^4}{4!}+\frac{(ir)^5}{5!}+\frac{(ir)^6}{6!}+\dots\\[2ex] &= \biggl( 1-\frac{r^2}{2!}+\frac{r^4}{4!}-\frac{r^6}{6!}+\dots \biggr) +i\,\biggl( \frac{r}{1!}-\frac{r^3}{3!}+\frac{r^5}{5!}-\dots \biggr)\\[3ex] &=\cos r+i\sin r \end{align} (of course he didn't write $n!$, which was invented later on).

Thus Euler derived that $\log(-1)=i\pi$, since $$ e^{i\pi}=\cos\pi+i\sin\pi=-1 $$

Euler's reasoning was rarely as rigorous as modern standards would require. He had a somewhat blind faith in how series work and often extended reasonings valid for polynomials to power series. And, sometimes, he made wrong assertions, which doesn't diminish at all his reputation as one of the greatest mathematicians of all time.

Decades later, Euler's reasoning could be placed on the solid bases of complex analysis. You can find something about this in Morris Kline's Mathematical Thought From Ancient to Modern Times, Volume I.

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It comes from the necessity to extend real exponentiation to the complex plane, so as to obtain a holomorphic function. So take real exponentiation series expansion $$e^x=\displaystyle\sum_{n=0}^{+\infty}\frac{x^n}{n!}$$ and put complex variable $z$ in place of $x$. What you get is an entire function (i.e holomorphic in the whole plane) which trivially coincides with real exponentiation when restricted to the real axis. Moreover, it is unique: another holomorphic extension of $e^x$ would coincide with it on the real axis, so that it would be the same function by identity principle.

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If you mean the rule that for $t\in \mathbb R$, one has that $e^{it}=\cos t+i\sin t $, then it follows from the Taylor expansion for $e^x=\sum_{k=1}^\infty\frac{x^k}{k!}$. It is a very nice exercise that when you plug into this expansion $x=it$, then upon splitting the series into its real and imaginary parts, one gets exactly the Taylor expansion for (respectively) $\cos t$ and $\sin t$.

Then, wishing to preserve the familiar properties of exponentiation, one obtains that for a general $z=x+iy$, $e^z=e^{x+iy}=e^xe^{iy}=e^x\cos y+i\cdot e^x\sin y$. Then, general complex exponentiation $z^w$ is obtained from the formula $z^w=e^{w\ln z}$, where $\ln$ is the multivalued function which is the inverse of $e^z$.

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A related question to ask is "what is exponentiation"? Several famous books, such as Rudin, start by defining $e^z$ as a shorthand for the following series: $$ e^z := \sum_n \frac{z^n}{n!} $$

From this approach, everything we are used to about exponentiation is derived from this taylor series.

For instance, substitute in a complex number and you have your answer.

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We wish to extend the definition of $e^x$ sa that it becomes meaningful when $x$ is replaced by any complex number $z$. we wish this extension to be such that that the law of exponents , $e^a.e^b=e^{a+b}$ , will be valid for all complex $a$ and $b$ ; and , of course , we would want $e^z$ to agree with the usual exponential when $z$ is real.If we write $z=x+iy$ , then , if the law of exponents is to be valid for complex numbers, then we must have $e^z=e^{x+iy}=e^x.e^{iy}$ , since $e^x$ is already known when $x$ is real , our task is to arrive at a reasonable definition at $e^{iy}$ when $y$ is real. Now , if $e^{iy}$ is to be a complex number , we may write $e^{iy}= A(y)+iB(y)$ ... (i) , where $A$ and $B$ are real-valued functions of $y$ to be determined. . Let us differentiate both sides , assuming $A$ and $B$ are differentiable and treating $i$ as a constant , we get $ie^{iy}=A'(y)+iB'(y)$ ...(ii), differentiatng once more , we find that $i.ie^{iy}=-e^{iy}=A''(y)+iB''(y)$ ... (iii) , comparison of (i) and (iii) shows that $A''(y)-A(y)$ and $B''(y)= -B(y)$ , in other words each of the functions $A$ and $B$ is a solution of the differential equation $f''+f=0$ , and from real analysis we know that this equation has exactly one solution with specified initial values $f(0)$ and $f'(0)$ . If we put $y=0$ in (i) and (ii) and use the fact that $e^0=1$ , we find that $A$ and $B$ have the initial values $A(0)=1 , A'(0)=0 $ and $B(0)=0 , B'(0)=1$ , hence $A$ and $B$ must be the cosine and sine functions respectively i.e.

$A(y) =$ cos $y$ and $B(y)=$ sin $y$ , in other words $e^{iy}=$ cos $y$ $+i$ sin $y$.

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As already noted, there are many ways to approach the complex exponential function. One additional way that hasn't been mentioned yet is to search for holomorphic solutions to the functional equation $$ f(z+w) = f(z)f(w).\tag{*} $$

If you normalize by saying $f'(0) = 1$, then $f(z) = e^z = e^x(\cos y + i\sin y)$ is the unique solution to (*).

Alternatively, you can define $e^z$ as the unique holomorphic solution to the differential equation $f' = f$ that satisfies $f(0) = 1$.

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