Let $$z_1$$ and $$z_2$$ be two complex numbers such that $$\bar{z}_1 = iz_2$$ and $$\arg\frac{z_1}{z_2} = \pi$$, then the argument of $$z_1$$ is

We express $$z_1$$ and $$z_2$$ in polar form as $$z_1 = re^{i\theta_1}$$ and $$z_2 = se^{i\theta_2}$$, and since $$\bar{z}_1 = iz_2$$, it follows that $$re^{-i\theta_1} = is\cdot e^{i\theta_2} = se^{i(\theta_2 + \pi/2)}$$.

Comparing moduli gives $$r = s$$, while comparing arguments yields $$-\theta_1 = \theta_2 + \frac{\pi}{2}$$, so that $$\theta_2 = -\theta_1 - \frac{\pi}{2}$$.

Next, using the condition $$\arg\!\bigl(\tfrac{z_1}{z_2}\bigr) = \pi$$, we have $$\arg\!\bigl(\tfrac{z_1}{z_2}\bigr) = \theta_1 - \theta_2 = \theta_1 - \bigl(-\theta_1 - \tfrac{\pi}{2}\bigr) = 2\theta_1 + \tfrac{\pi}{2} = \pi,$$ which implies $$2\theta_1 = \frac{\pi}{2}$$ and hence $$\theta_1 = \frac{\pi}{4}$$.

Therefore, $$\arg(z_1) = \frac{\pi}{4}$$, corresponding to Option C.