Partial derivative of the real part of a function

Question

I'm trying to understand the mathematical reasoning behind the example provided in this question.

If we have $$z = Ae^{i(\omega _{o}t+\phi )}$$

and define

$$x = Re (z),$$

then why is it that

$$\frac{\partial x}{\partial t} = Re \frac{\partial z}{\partial t},$$

instead of

$$\frac{\partial x}{\partial t} = \frac{\partial [Re(z)]}{\partial t}$$

In other words, why is it that the partial derivative of the real part of a function is equal to the real part of the partial derivative of the function, rather than being equal to the partial derivative of the real part of the function?

I'm just trying to understand the mathematical reasoning behind this, rather than why it is nonsensical from a physical standpoint.

Answer 1

It's due to the linearity of the derivative, if $t \in \mathbb{R}, z(t) = \sigma(t) + i \nu(t)$ then \begin{align*} \frac{\partial \text{Re}(z)}{\partial t} = \frac{\partial \text{Re}(\sigma + i \nu)}{\partial t} = \frac{\partial \sigma}{\partial t} = \text{Re}\left(\frac{\partial \sigma}{\partial t} + i \frac{\partial \nu}{\partial t}\right) = \text{Re}\left(\frac{\partial z}{\partial t}\right). \end{align*}

Answer 2

Although we can do this in general, let's look at your specific case. Note that Re$z=A\cos(\omega_0 t+\phi),$ so $\partial_t$Re$z=-A\omega_0\sin(\omega_0 t+\phi).$ On the other hand, $$\partial_t z=i\omega_0 z=-A\omega_0\sin(\omega_0 t+\phi)+iA\omega_0\cos(\omega_0 t+\phi),$$ and so Re($\partial_t z)=-A\omega_0 \sin(\omega_0 t+\phi).$ Hence, they're the same.

Answer 3

Use $z=A\cos(\omega_0 t + \phi) +i A\sin(\omega_0t + \phi)$ and take the derivative of this with respect to $t$ and you'll get $-A \omega_0 \sin(\omega_0 t + \phi) +iA\omega_0 \cos(\omega_0t + \phi)$. The real part of this is $-A \omega_0 \sin(\omega_0 t + \phi)$ which is the same thing as $\frac{\partial x}{\partial t}$