## Editorial for Inaho VI

**only**when stuck, and

**not to copy-paste code from it**. Please be respectful to the problem author and editorialist.

**Submitting an official solution before solving the problem yourself is a bannable offence.**

Author:

Without prior knowledge on the basics of complex numbers, this problem is impossible to solve. For example, it must be known that \(i = \sqrt{-1}\) and that the rectangular coordinate form of a complex number is \(z = a + bi\).

For the first subtask, since \(x \le 1\), we can utilize a simple if statement. If \(x = 0\), the solution is \(1 + 0i\). If \(x = 1\), the solution is simply \(0 + 1i\).

For the second subtask, some quick Googling will give the solutions to when \(x = 2\) and when \(x = 3\). When \(x = 2\), the solution is \(e^{-\frac{\pi}{2}} \approx 0.20788 + 0i\). When \(x = 3\), the solution is \(e^{\frac{\pi i e^{-\frac{\pi}{2}}}{2}} \approx 0.94716 + 0.32076i\). Alternatively, one can solve for when \(x = 2\) and \(x = 3\) manually using Euler's formula.

For the third subtask, one can utilize a for loop that iterates from \(1\) to \(x\). Start with \(2\) floating-point variables, \(a_0\) set to \(1\), and \(b_0\) set to \(0\). At each iteration of the for loop, it can be found that:

\(a_i = e^{-\frac{\pi b_{i-1}}{2}}\ cos \frac{\pi a_{i-1}}{2}\) and \(b_i = e^{-\frac{\pi b_{i-1}}{2}}\ sin \frac{\pi a_{i-1}}{2}\).

Alternatively, one can utilize a complex numbers library in their preferred programming language.

For the last subtask, one can figure out that solution converges to approximately \(0.4383 + 0.3606i\), which means that after a certain \(x\), the solution becomes consistent up to \(4\) decimal places. This means that we can hard code a value instead of looping up to \(x\). This certain \(x\) and the proof is left as an exercise for the reader.

**Time Complexity: \(O(x)\)**

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