Derivatives in time and phasor domains do not match

Question

Given that voltage can be represented in the time domain as the following:

$$V(t) = \vec{\mathbf{|V|}}\cos{(\omega t + \phi_V)}$$

It can be shown that in the phasor domain, this is the equivalent expression for V(t):

$$V(t) = \frac{1}{2}\vec{\mathbf{V}}e^{j\omega t} + \frac{1}{2}\vec{\mathbf{V^*}}e^{-j\omega t}$$

The time derivative in the first case is:

$$V'(t) = \vec{\mathbf{|V|}}(-w)\sin{(\omega t + \phi_V)}$$

My solution for the time derivative in the second case is:

$$V'(t) = \frac{1}{2}\vec{\mathbf{V}}(j \omega)e^{j\omega t} + \frac{1}{2}\vec{\mathbf{V^*}}(-j \omega)e^{-j\omega t}$$

$$V'(t) = (jw)(\frac{1}{2}\vec{\mathbf{V}}e^{j\omega t} - \frac{1}{2}\vec{\mathbf{V^*}}e^{-j\omega t})$$

Subtraction of the complex conjugates yields 2 times the imaginary part:

$$V'(t) = (jw)(2*Im{\frac{1}{2}\vec{\mathbf{V}}e^{j\omega t}})$$

$$V'(t) = \vec{\mathbf{|V|}}(jw)\sin{(\omega t + \phi_V)}$$

It appears that the derivatives vary by a factor of j. What is happening here?

Answer 1

It did not occur to me that taking the imaginary part would introduce a factor of j in the second to last step. Problem solved. Thanks.

Clarification edit: The difference of complex conjugates is 2j times the imaginary part, not just 2 times. This is because Re{a+bj} = a and Im{a+bj} = b. Then (a+bj)-(a-bj) = 2bj.