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math771
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Hi Peter,
I am also currently reading Needham's book and am at a similar point as you. From my understanding, the "moving particle argument" is a heuristic way of visualizing and understanding Euler's formula, but it is not a rigorous proof. The idea is that as the particle moves along the curve Z(t) = e^{it}, its velocity is always at a right angle to its position vector, which is due to the fact that the derivative of e^{it} is ie^{it}, causing a rotation of 90 degrees.
Now, as for why this results in the particle traveling around a unit circle, it is not immediately obvious. Needham states that it is clear, but as you pointed out, it is not rigorously proven. However, we can think of it in the following way: as the particle moves along the curve, its position vector is constantly changing, but its length remains constant at 1 (since e^{it} has magnitude 1). This means that the particle is always a distance of 1 from the origin, forming a circle of radius 1. Since the velocity is always at a right angle to the position vector, it will always be tangent to this circle, causing the particle to continuously travel around it.
As for the last paragraph, Needham is saying that after a time t = \theta, the particle will have traveled a distance of \theta around the unit circle, and its new position will be at an angle of \theta from the positive real axis. This is exactly what Euler's formula states: e^{i\theta} = cos(\theta) + isin(\theta). The real part represents the x-coordinate on the unit circle and the imaginary part represents the y-coordinate, forming a right triangle with a hypotenuse of length 1 (since the particle is always a distance of 1 from the origin).
I hope this helps clarify the moving particle argument and its connection to Euler's formula. Let me know if you have any further questions.
I am also currently reading Needham's book and am at a similar point as you. From my understanding, the "moving particle argument" is a heuristic way of visualizing and understanding Euler's formula, but it is not a rigorous proof. The idea is that as the particle moves along the curve Z(t) = e^{it}, its velocity is always at a right angle to its position vector, which is due to the fact that the derivative of e^{it} is ie^{it}, causing a rotation of 90 degrees.
Now, as for why this results in the particle traveling around a unit circle, it is not immediately obvious. Needham states that it is clear, but as you pointed out, it is not rigorously proven. However, we can think of it in the following way: as the particle moves along the curve, its position vector is constantly changing, but its length remains constant at 1 (since e^{it} has magnitude 1). This means that the particle is always a distance of 1 from the origin, forming a circle of radius 1. Since the velocity is always at a right angle to the position vector, it will always be tangent to this circle, causing the particle to continuously travel around it.
As for the last paragraph, Needham is saying that after a time t = \theta, the particle will have traveled a distance of \theta around the unit circle, and its new position will be at an angle of \theta from the positive real axis. This is exactly what Euler's formula states: e^{i\theta} = cos(\theta) + isin(\theta). The real part represents the x-coordinate on the unit circle and the imaginary part represents the y-coordinate, forming a right triangle with a hypotenuse of length 1 (since the particle is always a distance of 1 from the origin).
I hope this helps clarify the moving particle argument and its connection to Euler's formula. Let me know if you have any further questions.