Rectangular spiral

Using complex number $z=x+yi$ . we note that $z_1 = 1$ , $z_2 = 1 + \dfrac i2$ , $z_3 = 1 + \dfrac i2 + \dfrac {i^2}3$ , implies that:

$\begin{aligned} z_\infty & = 1 + \frac i2 + \frac {i^2}3 + \dfrac {i^3}4 + \cdots \\ & = \frac 1i \left(i + \frac {i^2}2 + \frac {i^3}3 + \dfrac {i^4}4 + \cdots \right) & \small \color{#3D99F6} \text{Note that } - \ln (1-x) = \sum_{k=1}^\infty \frac {x^k}k \\ & = - \frac 1i \ln (1-i) \\ & = i \ln \left( \sqrt 2 \left(\frac 1{\sqrt 2}- \frac i{\sqrt 2}\right)\right) & \small \color{#3D99F6} \text{By Euler's formula }e^\theta = \cos \theta + i\sin \theta \\ & = i \ln \left( \sqrt 2 e^{-\frac \pi 4i}\right) \\ & = i \left( \ln \sqrt 2 - \frac \pi 4i\right) \\ & = \frac \pi 4 + i \ln \sqrt 2 \end{aligned}$

$\implies x = \dfrac \pi 4$ code: 4; $y = \ln \sqrt 2$ code: 2. Therefore, the answer is $\boxed{42}$ .

We have $x = 1 - \frac13 + \frac15 - \frac17 +-\cdots \\ = \frac{1^1}1 - \frac{1^3}3 + \frac{1^5}5 - \frac{1^7}7 +- \cdots \\ = \text{arc tg}\ 1 = \frac \pi 4\approx 0.7854;$ and $y = \frac12 - \frac14 + \frac16 - \frac 18 +-\cdots \\ = \frac12\left(\frac {1^1}1 - \frac{1^2}2 + \frac{1^3}3 - \frac{1^4}4 +-\cdots\right) \\ = \frac12\cdot \ln (1+1) = \frac12\ln 2 = \ln\sqrt 2 \approx 0.3466.$ Thus $x$ is a rational multiple of $\pi$ (option 4) and $y$ is the natural logarithm of the square root of a rational number (option 2). The solution is $\boxed{42}$ .