operatorname{Arg}left(frac{4+2 i}{1-2 i}+frac | vedclass

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Question

(A) First,simplify the expression inside the argument: $\frac{4+2 i}{1-2 i} = \frac{(4+2 i)(1+2 i)}{(1-2 i)(1+2 i)} = \frac{4+8 i+2 i-4}{1+4} = \frac{

Vedclass · Accepted Answer

$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

Vedclass · Answer

(A) First,simplify the expression inside the argument: $\frac{4+2 i}{1-2 i} = \frac{(4+2 i)(1+2 i)}{(1-2 i)(1+2 i)} = \frac{4+8 i+2 i-4}{1+4} = \frac{10 i}{5} = 2 i$ $\frac{3+4 i}{2+3 i} = \frac{(3+4 i)(2-3 i)}{(2+3 i)(2-3 i)} = \frac{6-9 i+8 i+12}{4+9} = \frac{18-i}{13}$ Now,add the two results: $2 i + \frac{18-i}{13} = \frac{26 i + 18 - i}{13} = \frac{18+25 i}{13} = \frac{18}{13} + \frac{25}{13} i$ The argument is $	an^{-1}\left(\frac{25/13}{18/13}ight) = 	an^{-1}\left(\frac{25}{18}ight)$. Since $\frac{25}{18} > 1$,we have $	an^{-1}\left(\frac{25}{18}ight) > 	an^{-1}(1) = \frac{\pi}{4}$. Also,since $\frac{25}{18} > 0$,the argument lies in the first quadrant,i.e.,$(0, \frac{\pi}{2})$. Thus,the value lies in the interval $\left(\frac{\pi}{4}, \frac{\pi}{2}ight)$.

$\operatorname{Arg}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)$ lies in the interval

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