Question

The principal argument of $\frac{i-3}{i-1}$ is

• A. ${\tan }^{-1}\frac{1}{2}$
• B. ${\tan }^{-1}\frac{3}{2}$
• C. ${\tan }^{-1}\frac{5}{2}$
• D. ${\tan }^{-1}\frac{7}{2}$

Answer 1

The correct option is A ${\tan }^{-1}\frac{1}{2}$

We have,

arg$(i-3)=ta{n}^{-1}\left(\frac{-1}{3}\right)$ &

arg$(i-1)=ta{n}^{-1}(-1)$

arg$(i-3)-$arg $(i-1)=ta{n}^{-1}\left(\frac{\frac{-1}{3}+1}{1+\frac{1}{3}}\right)=ta{n}^{-1}\left(\frac{1}{2}\right)$