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Converting between polar and rectangular coordinates

Problem 8.149 · easy

Convert the rectangular point \( \displaystyle \left(2, - 2 \sqrt{3}\right) \) to polar coordinates with \( \displaystyle r > 0 \) and \( \displaystyle 0 \le \theta < 2\pi \).

\[ 4 \]
r = √(x² + y²).✓ Proved
The point is in quadrant 4; θ = 5*pi/3.
Reviewed
\[ \left[\begin{matrix}2\\- 2 \sqrt{3}\end{matrix}\right] \]
Converting back gives the point.✓ Proved

Answer \( \left(4, \frac{5 \pi}{3}\right) \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	r and θ from hypot and atan2

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the quadrant and calculates the angle. The steps are logically sound and the final answer matches the problem constraints.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-06
qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly identifies the quadrant and calculates the angle. The steps are logically sound and the final answer matches the problem constraints.
qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly identifies the quadrant and calculates the angle. The verification step confirms the conversion is correct.
gpt-oss:20b: pass 2026-10-06

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/polar_conversion, checked 2026-10-06 with SymPy 1.14.0.