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I recently repeated some math basics of the Fourier transform and of course stumbled across Euler's formula. When reading the term $\cos(x) + i\sin(x)$ I wondered why it could not be written as $2\cos(x)$. Since all professors always emphasize that a cosine is nothing but a $90$ degree shifted sine, I was wondering why the multiplication with i, which also causes a $90°$ shift on the complex plane, doesn't result in a $\cos$-function.

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    $\begingroup$ note that for any chosen $x\in\Bbb R$ we have that $i\sin(x)$ is a purely imaginary number, however $2\cos (x)$ is real, so they cannot be equal $\endgroup$ – Masacroso May 1 at 10:49
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"Cosine is 90 degree shifted sine" means a shift in the arguments, not the rotation of the values on the complex plane. That is $ \cos(x) = \sin(x+90^\circ) $. Multiplication by $i$ is rotating the value of the function, not shifting the argument.

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  • $\begingroup$ I encountered another problem: If it is like you say, why is it that differentiation in the time domain (which would make a sine a cosine) is multiplication with i in the frequency domain? This is why the impedances of capacitors and inductors have the i in their formula, isnt it? But if cos(x) is not i*sin(x), how does this make sense? $\endgroup$ – Philipp317 Jul 23 at 6:38
  • $\begingroup$ @Philipp317 I would need more contest to be sure, but I think you're talking about a case of alternating current, where for example we assume that the charge on the capacitor is $Q(t) = {\rm Re}(Q_0 e^{i\omega t})$. Then you have $$ I = \frac{dQ}{dt} = {\rm Re}(i\omega Q_0 e^{i\omega t}) = {\rm Re}(I_0 e^{i\omega t})$$ where $I_0 = i\omega Q_0$. $\endgroup$ – Adam Latosiński Jul 23 at 9:06
  • $\begingroup$ I was talking about the current-voltage relation for a capacitor. It's described by I = C * dU/dt , basically meaning that if the applied voltage is a sine-wave, the resulting current is a cosine-wave. It effectively means a 90° degree shift for a harmonic input. The expression to describe the impedance of a capacitor in the frequency domain is 1/jwC , and here I thought the division by j in the denominator expresses the same phase shift the differential equation in the time domain described. But the answers told me, those 2 "phase shifts" are not the same $\endgroup$ – Philipp317 Jul 24 at 15:39
  • $\begingroup$ Division by $i$ is a phase shift in the oposite direction because while $i=e^{i\pi/2}$, we have $\frac{1}{i} = -i = e^{-i\pi/2 }$. $\endgroup$ – Adam Latosiński Jul 24 at 17:59
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    $\begingroup$ No, you must compare real part to the real part and imaginary part to the imaginary part. That gives you $$ -sin(\omega t) = \cos(\omega t + \pi/2)$$ $$ \cos(\omega t) = \sin(\omega t+\pi/2)$$ $\endgroup$ – Adam Latosiński Jul 28 at 7:33
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It causes a shift when we deal with angles and not with functions of angles.

For example, in a frequency domain, we may write, $5i$ as $5\angle90^o$, which is a notation for $5e^{(i\omega t +\pi/2)}$

But here, we have $i. sin(x)$ and not a magnitude of a time or frequency domain voltage or current or whatsoever.

So, $i\cdot\sin(x)$ is not equal to $cos(x)$ . Also only $\sin(90°\pm x) = \cos(x)$, not when we have $i\cdot \sin(x)$.

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