Le 26/05/2025 à 22:30, Richard Hachel a écrit :

Z=r(cosθ+i.sinθ), I understand.

Are you sure of that? I doubt it.

Z=e^iθ, I don't understand the expansion.

No doubt about that. What you wrote is incorrect, the correct expression
is Z = r*e^(iθ)

Nice that you asked though.

Does anyone understand?

Sure.

sin(x) = sum_(k=0)^\infty ((-1)^k x^(1 + 2 k))/((1 + 2 k)!)

cos(x) = sum_(k=0)^\infty ((-1)^k x^(2 k))/((2 k)!)

e^x = sum_(k=0)^∞ x^k/(k!)

i^2 = -1

Hence:

e^(ix) = cos(x) + i*sin(x)

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Z=r(cosθ+i.sinθ), I understand.
Z=e^iθ, I don't understand the expansion.
Does anyone understand?

R.H.

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Op 26/05/2025 om 22:30 schreef Richard Hachel:

Z=r(cosθ+i.sinθ), I understand.
Z=e^iθ, I don't understand the expansion.
Does anyone understand?

R.H.

Some people claim that you're not supposed to understand math.
All you can hope for is to get used to it.

I think John Von Neumann said something along those lines.

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Op 26/05/2025 om 22:56 schreef Python:

Le 26/05/2025 à 22:30, Richard Hachel a écrit :

Z=r(cosθ+i.sinθ), I understand.

Are you sure of that? I doubt it.

Z=e^iθ, I don't understand the expansion.

No doubt about that. What you wrote is incorrect, the correct expression
is Z = r*e^(iθ)

Nice that you asked though.

Does anyone understand?

Sure.

sin(x) = sum_(k=0)^\infty ((-1)^k x^(1 + 2 k))/((1 + 2 k)!)

cos(x) = sum_(k=0)^\infty ((-1)^k x^(2 k))/((2 k)!)

e^x = sum_(k=0)^∞ x^k/(k!)

i^2 = -1

Hence:

e^(ix) = cos(x) + i*sin(x)

https://youtu.be/FFPXm-tuOt8?t=263

We can kind of get a feeling for it by interactively observing the
Taylor expansion (using the slider for N) to see how they approach the transcendental functions with polynomials.

https://www.desmos.com/calculator/z2jjsuv5o2

https://youtu.be/3geVAJvJM8c?t=701

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