function $(-2)^{x}$

Question

What is the real and imaginary parts for the function $f(x)=(-2)^{x}$ ? Is there a unique solution to this question?

Answer 1

Since $-2 = 2e^{i\pi}$, then $\ln(-2) = \ln(2)+i(\pi+2k\pi) = \ln(2)+i(2k+1)\pi$ are the complex logarithm values of $\ln(-2)$.

Since $a^x = e^{x\ln(a)}$, we have $$\begin{align*} (-2)^x &= \exp(x\ln(-2))\\ &= \exp\left(x\Bigl( \ln(2) + i(2k+1)\pi\Bigr)\right)\\ &= \exp(x\ln(2) + ix(2k+1)\pi)\\ &= e^{x\ln(2)} e^{ix(2k+1)\pi}\\ &= 2^x\Bigl( \cos(x(2k+1)\pi) + i\sin(x(2k+1)\pi)\Bigr)\\ &= 2^x\cos(x(2k+1)\pi) + i2^x\sin(x(2k+1)\pi). \end{align*}$$ So for integer $k$, the different values of $(-2)^x$ have real part $2^x\cos\Bigl(x(2k+1)\pi\Bigr)$, and complex part $2^x\sin\Bigl(x(2k+1)\pi\Bigr)$.

Added. To take the principal value of the logarithm (which requires the imaginary part to lie in $(-\pi,\pi]$, then you use $k=0$.

Answer 2

Hint: by definition, $a^x = e^{x \log a}$. Complicating matters: there are infinitely many branches of the log, corresponding to branches of $a^x$.