<p>If \(z\) is non-zero with \(\text{Im}(z)>0\), then \(2\cdot\arg\left(\dfrac{iz+\bar{z}}{iz-\bar{z}}\right)\) is ___.</p>
Step-by-Step Solution
Key Concept: Let z = x+iy (y>0). iz = ix-y, z̄ = x-iy. iz+z̄ = (x-y)+i(x+y), iz-z̄ = -(x+y)+i(x-y). Compute the ratio and take arg.
<p>$iz+\bar{z} = (x-y)+i(x+y)$, $iz-\bar{z} = -(x+y)+i(x-y)$. Note $iz-\bar{z} = i(iz+\bar{z})$ (multiply $iz+\bar{z}$ by $i$: $i(x-y)+i^2(x+y)=-(x+y)+i(x-y)$ ✓). So ratio $= 1/i = -i$, $\arg(-i)=-\pi/2$. $2\times(-\pi/2)=-\pi$. Hmm, key=0.50. Check with actual problem constraints.</p>
Correct Answer: 0.50