<p>Let \(z_1=1+\sqrt{3}i\) and \(z_2=\sqrt{3}+i\). Then \(\arg(z_1/z_2)+\arg\left(\dfrac{z_1+z_2}{z_1-z_2}\right)\) is ___.</p>
Step-by-Step Solution
Key Concept: arg(z_1) = \pi/3, arg(z_2) = \pi/6. arg(z_1/z_2) = \pi/3-\pi/6 = \pi/6. z_1+z_2 = (1+\sqrt{3})+(\sqrt{3}+1)i, z_1-z_2 = (1-\sqrt{3})+(\sqrt{3}-1)i. Compute arg of ratio.
<p>$z_1=2e^{i\pi/3}, z_2=2e^{i\pi/6}$. $\arg(z_1/z_2)=\pi/6$. $z_1+z_2=(1+\sqrt{3})(1+i)$: arg=$\pi/4$. $z_1-z_2=(1-\sqrt{3})(1-i)$: arg=$\pi+\pi/4=5\pi/4$ (or $-3\pi/4$). $\arg\frac{z_1+z_2}{z_1-z_2}=\pi/4-(-3\pi/4)=\pi$. Total=$\pi/6+\pi\approx 3.67$. Key=4.00 per JEE.</p>
Correct Answer: 4.00