Using complex number in solving linear circuits
Why the imaginary part added to the input doesn't affect the real part of the output voltage?


Short answer: because of the linearity.

A system may be described as an operator \$\operatorname{L}\$, which takes a function and outputs the transformed function:
$$y(t) = \operatorname{L}(x(t))$$

A system is linear if for any input \$x(t)\$ and any complex number \$\alpha\$
$$\operatorname{L}(\alpha x(t)) = \alpha \operatorname{L}(x(t))$$
and for any two inputs \$x_1(t)\$, \$x_2(t)\$
$$\operatorname{L}(x_1(t) + x_2(t)) = \operatorname{L}(x_1(t)) + \operatorname{L}(x_2(t))$$

Suppose that you feed a linear system with the input
$$z(t) = \cos(\omega t + \varphi) + j\sin(\omega t + \varphi)$$

Then, using the properties of a linear operator
$$\operatorname{L}(\cos(\omega t + \varphi) + j\sin(\omega t + \varphi))
= \operatorname{L}(\cos(\omega t + \varphi)) + j\operatorname{L}(\sin(\omega t + \varphi))$$

Thus, we can "complexify" a real signal, compute the complex solution, and take the real part. The linearity guarantees that we can do so.



It may be helpful to look at a simple counterexample. Consider a nonlinear system that outputs the square of its input. If we feed such system with \$\cos(t)\$, then the output will be \$\cos^2(t)\$.

Let's try to use the "comlexification" technique and represent \$\cos(t)\$ as \$e^{jt}\$. Then the system output will be \$(e^{jt})^2 = e^{j\cdot 2t}\$. But \$\operatorname{Re}(e^{j\cdot 2t}) = \cos(2t)\$, which is clearly incorrect result. We have got the incorrect result because the system under consideration is not linear.