Periodic BC's of heat equation

[Panphobia]

Homework Statement

I have the heat equation
$$u_t=u_{xx}$$
$$u(0,t)=0$$
$$u(1,t) = \cos(\omega t)$$
$$u(x,0)=f(x)$$
Find the stable state solution.

The Attempt at a Solution

I used a transformation to complex to solve this problem, and then I can just take the real part to the complex solution to solve this problem. I want a steady state periodic solution. So the new problem is
$$v_t=v_{xx}$$
$$v(0,t)=0$$
$$v(1,t) = e^{i\omega t}$$
$$v(x,0)=f(x)$$
Using separation of variables, solving for T(t) and X(x) we get $$T(t) = Ce^{i\omega t}$$ and $$X(x) = Ae^{\sqrt{\frac{\omega}{2}} (1+ i)x} +Be^{-\sqrt{\frac{\omega}{2}} (1+ i)x}$$ but the first BC gives us $$B=-A$$, and without applying the second BC we get

$$v(x,t) = X(x)T(t) = A\bigg(e^{\sqrt{\frac{\omega}{2}} (1+ i)x} -e^{-\sqrt{\frac{\omega}{2}} (1+ i)x}\bigg)e^{i \omega t}$$

Now if I apply the second BC $$v(1,t)=e^{i\omega t}$$, I get $$ A =\frac{1}{e^{\sqrt{\frac{\omega}{2}} (1+ i)} -e^{-\sqrt{\frac{\omega}{2}} (1+ i)}}$$. Then I can't separate out a real solution because the denominator is split up into a real and complex solution. Did I go wrong somewhere in my steps? I know I can solve it a different way, but the question in the textbook told me to do it this way.

 

[Orodruin]

The following identity may be of interest:
$$
\frac{1}{z} = \frac{z^*}{zz^*} = \frac{z^*}{|z|^2}
$$

 

[Orodruin]

Also, note that ”periodic boundary conditions” would typically be used to refer to problems that are periodic in x, such as the heat equation on a circle.