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The inspiration for this series was I tried to prove the binomial theorem for complex numbers(which I eventually did) but on one attempt ran into an interesting series. Im curious about whether the series is valid, because I was able to prove the theorem works using it. It starts like this, let $z = a + bi$ \begin{align} (1+x)^z = (1+x)^{a+bi} = (e^{\ln(1+x)})^{a+bi} = e^{a \ln (1+x)} \cdot e^{bi\ln (1 + x)} \end{align} Note that $x > -1$. I could simplify the second part with Euler's formula, but instead used the Taylor series of $e^x$(which appears to work with complex numbers because thats how Euler's formula can be derrived). Let $u = x + 1$, it follows that \begin{align} (1+x)^z & = (1 + a \ln u + \frac{a^2 \ln^2 u}{2} + \frac{a^3 \ln^3 u}{3!} + \ldots)\cdot(1 + bi \ln u - \frac{b^2 \ln^2 u}{2} - \frac{ib^3 \ln^3 u}{3!} + \ldots) \end{align} Recall that: \begin{align} (a + bi)^1 &= a + bi\\ (a + bi)^2 &= a^2 + 2abi - b^2\\ (a + bi)^3 &= a^3 + 3a^2bi - 3ab^2 - ib^3 \end{align} Let set $X$ contain the terms of the first sum, and $Y$ the terms of the second. It can be seen that these powers of $z$ can be constructed from the infinite sum product as such \begin{equation} \frac{(a + bi)^n \ln^n u}{n!} = X_{n+1}Y_{1} + X_n Y_2 + \ldots + X_2 Y_n + X_1 Y_{n+1} \end{equation} e.g 4th power \begin{align} & X_5 Y_1 + X_4 Y_2 + X_3 Y_3 + X_2 Y_4 + X_1 Y_5\\ &= \frac{a^4 \ln^4 u}{4!} \cdot 1 + \frac{a^3 \ln^3 u}{3!} \cdot bi \ln u + \frac{a^2 \ln^2 i}{2} \cdot \frac{-b^2 \ln^2 u}{2} + a \ln u \cdot \frac{-b^3i \ln^3 u}{3!} + 1 \cdot \frac{b^4 \ln^4 u}{4!}\\ &= \ln^4 u(\frac{a^4}{4!} + \frac{4a^3 bi}{4 \cdot 3!} - \frac{3! \cdot a^2b^2}{4 \cdot 3!} -\frac{4 \cdot ab^3i}{4 \cdot 3!} + \frac{b^4}{4!})\\ &= \frac{ln^4 u}{4!}(a^4 + a^3bi - a^2b^2 - ab^3i +b^4) = \frac{(a+bi)^4 \ln^4 u}{4!} \end{align} Since each pair of the $X$ and $Y$ elements appears uniquely, and all the pairs will be present in the multiplication between the two infinite sums, it follows that \begin{equation} (1 + x)^z = 1 + z\ln(1+x) + \frac{z^2 \ln^2(1+x)}{2!} + \frac{z^3 \ln^3(1+x)} {3!} + \ldots, \text{$x > - 1$} \end{equation} Is this sum correct? and if it is not, where is the error in my reasoning?

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Yes, this is correct: what you have essentially done is derived the expansion of $(1+x)^{z} := \exp{(z\log{(1+x)})}$ in powers of $\log{(1+x)}$. It converges everywhere that $\log{(1+x)}$ is defined, since the power series of the exponential converges for every complex number. Hence you have a definition of $(1+x)^z$ that works for $x$ anywhere the whole complex plane apart from the real axis with $x<-1$.

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