Natural Log of a Negative Number

$\begin{aligned} \ln(-3)&=\ln(3\times-1)\\ &=\ln(3)+\ln(-1)\\ &=\ln(3)+\ln(e^{i(2k+1)\pi})\\ &=\ln(3)+i(2k+1)\pi\\ &\Leftrightarrow\ln(A)+i(2k+1)B\\ &\Rightarrow (A,B)=(3,\pi) \end{aligned}$

$\begin{aligned} ⌊1000S_1⌋+⌊100S_2⌋&=\left⌊1000\left(\frac\pi3\right)\right⌋+⌊100\pi(3)(\pi)⌋\\ &=\left⌊\frac{1000}{3}\pi\right⌋+\left⌊300\pi^2\right⌋\\ &=⌊1047.1\dots⌋+⌊2960.8\dots⌋\\ &=1047+2960\\ &=\boxed{4007} \end{aligned}$

$\begin{aligned} z & = \ln (-3) = \ln (3) + \ln({\color{#3D99F6}-1}) & \small \color{#3D99F6} \text{By Euler's formula: }e^{i \theta} = \cos \theta + i\sin \theta \\ & = \ln (3) + \ln \left(\color{#3D99F6} e^{(2k+1)i \pi}\right) & \small \color{#3D99F6} \text{where }k \text{ is an integer.} \\ & = \ln (3) + i (2k+1)\pi \end{aligned}$

Therefore $A=3$ and $B=\pi$ . Then $S_1 = \dfrac \pi 3 \approx 1.047197551$ and $S_2 = \pi (3)(\pi) \approx 29.6088132$ and $\lfloor 1000S_1 \rfloor + \lfloor 100S_2 \rfloor = 1047+2960 = \boxed {4007}$ .

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Reference: Euler's formula

Main steps of the problem: $z = \ln(-3)$ $z = \ln(3) + \ln(-1)$ Since $i^2 = -1$ $z = \ln(3) + \ln(i^2)$ Also, $i^2 = e^{(2k+1)\pi}$ , which implies: $\ln(i^2) = (2k+1)\pi$ , where $k$ belongs to the set of integers. This leads to: $z = \ln(3) + (2k+1)\pi$

Which implies $A=3$ and $B = \pi$

From here the answer can be found by computing: $S_1 = \frac{B}{A}$ $S_2 = \pi AB$

The Slope of the line $S_1 = \frac{B}{A} = 1.047197$

The area of the ellipse is $S_2 = \pi AB = 29.6088$

From here: $X = 1000S_1 = 1047.97$ and $Y = 100S_2 = 2960.88$

Finally, $\boxed{\lfloor X \rfloor + \lfloor Y \rfloor = 4007}$