Complex Power of a Complex Number Using Euler's Formula

Question

To get the principal answer to a complex number $(a+bi)$ raised to another complex number $(c+di)$ I understand you can get this by first determining $r=\sqrt{a^2+b^2}$ and $\theta = \arctan(b/a)$ and then use those by raising $e^{(\ln(r)*c-d*\theta)+(\ln(r)*d+c*\theta)}$. But I find that does not provide the correct answer for an equation like $(-1+1.732i)^{3}$

Here $r=\sqrt{(-1)^2+1.732^2} = 2$ And $\theta = \arctan{\frac{1.732}{-1}} = -1.04718$

So plugging that in we get $e^{(\ln{2}*3-0*(-1.04718))+(\ln{2}*0+3*(-1.04718))} = -8$

But if I expand the equation at manually (or check it with Wolfram Alpha) I find the correct answer is actually $+8$

It seems there must be a correction factor needed sometimes to switch the sign of the result? I thought it might be related to odd number exponents (in this case $3$) but it doesn't seem to be needed all the time (for example $(2+0i)^3$ works just fine without a correction needed.

I've spend some time researching and I can't see this addressed anywhere I've looked in regard to Euler's Formula or the De Moivre theorem. Does anyone know the answer as to how to fix this?

Later edit: I arbitrarily changed the program I was working on to switch the sign of the answer produced by the above formula when: the real part of the exponent was odd AND the real part of the base was less than zero. So far, my revised program has correctly answered the dozen or so questions I have given it.

But I don't know if that is just a fluke. And even if it is correct to reverse the sign of the answer like that I don't know WHY it has to be done.

So I continue to look for some wisdom on this.

Answer 1

Take $1.732$ = $\sqrt 3 $ just for simplification . The mistake you did is that you took $\theta = \arctan(\frac{y}{x})$, which is true only when the point lies in the 1st quadrant in the complex plane. But in the question the point in the base $(-1,1.732)$ lies in 2nd quadrant
So its argument will be $\pi- \arctan(|\frac{y}{x}|) $which gives the value $2\pi/3$. Now putting the respective values in the formula you mentioned We $(e^ {3\log2} ). e^ {(2\pi/3)×3}$. Which gives us $+8.$

Answer 2

Your calculation of arc-tan took the wrong branch of the tangent.

Note that in your number $(−1+1.732\,i)$ the real part is negative and the imaginary part is positive, so the number belongs to the second quadrant and the actual argument is between $\pi/2$ and $\pi$.
Actually, it is $\theta = \frac 23\pi \approx +2.0944,$ which is $\left(\arctan \frac {1.732}{-1}\right) + \pi,$
whilst your result of $−1.04718 \approx \arctan \frac{-1.732}1$ corresponds to $(1-1.732\,i)$ in the fourth quadrant.