Assume we have a purely inductive circuit such as below:

To calculate the most general case, \$V_1\$ is given as such:

$$v(t) = V_m \cdot cos(\omega t+q)$$

For an inductor, \$V = L \frac{di}{dt}\$. Writing it in integral form:

$$i(t=0)+\int_0^t v \,dt= i(t)$$

Now consider \$i(t=0)=0\$, meaning there was no current in the system before the source was on:

$$\int_0^t V_m\cdot cos(\omega t+q) \,dt= i(t)$$

Moreover, the result of the integral, and hence the current:

$$\frac{V_m}{\omega}\cdot(sin(\omega t+q)-sin(q))= i(t)$$

If transformed to their complex counterparts via \$e^{j\omega t} = cos(\omega t)+jsin(\omega t)\$, the result, which is the impedance, is not \$j\omega L\$ but some term similar to \$j\omega L\$, but also containing the phase of the voltage source, which shouldn't be there, since impedance is not dependent on the phase of the source.

Where am I going wrong?

Assume we have a purely inductive circuit such as below:

To calculate the most general case, \$V_1\$ is given as such:

$$v(t) = V_m \cdot \cos(\omega t+\theta)$$

For an inductor, \$v(t) = L \dfrac{\mathrm{d}i(t)}{\mathrm{d}t}\$. Writing it in integral form:

$$i(0)+\int_0^t{v(t)\mathrm{d}t}= i(t)$$

Now consider \$i(0)=0\$, meaning there was no current in the system before the source was on:

$$\int_0^t{V_m\cdot \cos(\omega t+\theta) \mathrm{d}t}= i(t)$$

Moreover, the result of the integral, and hence the current:

$$\dfrac{V_m}{\omega}\cdot(\sin(\omega t+\theta)-\sin(\theta))= i(t)$$

If transformed to their complex counterparts via \$\mathrm{e}^{j\omega t} = \cos(\omega t)+j\sin(\omega t)\$, the result, which is the impedance, is not \$j\omega L\$ but some term similar to \$j\omega L\$, but also containing the phase of the voltage source, which shouldn't be there, since impedance is not dependent on the phase of the source.

Where am I going wrong?

assume we have purely inductive circuit such as below  

to calculate the most general case, V1 is given as such  : v(t)= Vm*cos(wt+q)

since , for inductor: v=L di/dt. Writing it in integral form , $$i(t=0)+\int_0^t v \,dt= i(t)$$

now consider, i(t=0)=0, meaning there was no current in the system before the source is on.
$$\int_0^t Vm*cos(wt+q) \,dt= i(t)$$ moreover, the result of the integral, and hence the current: $$(Vm/w)*(sin(wt+q)-sin(q))= i(t)$$

if transformed to their complex counterparts via "e^jwt=cos(wt)+jsin(wt)" , the result  , which is the impedance, is  not coming as $$ jwL$$ but some term similar "jwL" but also containing the phase of the voltage source, which shouldn't be since impedance is not dependent on the phase of the source. where am I doing the wrong

Assume we have a purely inductive circuit such as below:

To calculate the most general case, \$V_1\$ is given as such:

$$v(t) = V_m \cdot cos(\omega t+q)$$

For an inductor, \$V = L \frac{di}{dt}\$. Writing it in integral form:

$$i(t=0)+\int_0^t v \,dt= i(t)$$

Now consider \$i(t=0)=0\$, meaning there was no current in the system before the source was on:

$$\int_0^t V_m\cdot cos(\omega t+q) \,dt= i(t)$$

Moreover, the result of the integral, and hence the current:

$$\frac{V_m}{\omega}\cdot(sin(\omega t+q)-sin(q))= i(t)$$

If transformed to their complex counterparts via \$e^{j\omega t} = cos(\omega t)+jsin(\omega t)\$, the result, which is the impedance, is not \$j\omega L\$ but some term similar to \$j\omega L\$, but also containing the phase of the voltage source, which shouldn't be there, since impedance is not dependent on the phase of the source.

Where am I going wrong?

assume we have purely inductive circuit such as below

to calculate the most general case, V1 is given as such : v(t)= Vm*cos(wt+Qv)

since , for inductor: v=L di/dt. Writing it in integral form , $$i(t=0)+\int_0^t v \,dt= i(t)$$

now consider, i(t=0)=0, meaning there was no current in the system before the source is on.
$$\int_0^t Vm*cos(wt+q) \,dt= i(t)$$ moreover, the result of the integral, and hence the current: $$(Vm/w)*(sin(wt+q)-sin(q))= i(t)$$

if transformed to their complex counterparts via "e^jwt=cos(wt)+jsin(wt)" , the result , which is the impedance, is not coming as $$ jwL$$ but some term similar "jwL" but also containing the phase of the voltage source, which shouldn't be since impedance is not dependent on the phase of the source. where am I doing the wrong

assume we have purely inductive circuit such as below

to calculate the most general case, V1 is given as such : v(t)= Vm*cos(wt+q)

since , for inductor: v=L di/dt. Writing it in integral form , $$i(t=0)+\int_0^t v \,dt= i(t)$$

now consider, i(t=0)=0, meaning there was no current in the system before the source is on.
$$\int_0^t Vm*cos(wt+q) \,dt= i(t)$$ moreover, the result of the integral, and hence the current: $$(Vm/w)*(sin(wt+q)-sin(q))= i(t)$$

if transformed to their complex counterparts via "e^jwt=cos(wt)+jsin(wt)" , the result , which is the impedance, is not coming as $$ jwL$$ but some term similar "jwL" but also containing the phase of the voltage source, which shouldn't be since impedance is not dependent on the phase of the source. where am I doing the wrong