Question

If $ z=\sqrt{3}+i, $ then the argument of $ {{z}^{2}}{{e}^{z-i}} $ is equal to

• A. $ \frac{\pi }{2} $
• B. $ \frac{\pi }{6} $
• C. $ {{e}^{\pi /6}} $
• D. $ {{e}^{\pi /3}} $
• E. $ \frac{\pi }{3} $

Answer 1

Answer: (E)

Given, $ z=\sqrt{3}+i, $ $ \arg ({{z}^{2}}{{e}^{z-i}})=\arg [(3-1+2\sqrt{3}i){{e}^{\sqrt{3}}}] $
$=\arg [(2+2\sqrt{3}i){{e}^{\sqrt{3}}}] $
$={{\tan }^{-1}}\left[ \frac{2\sqrt{3}}{2} \right] $
$=\frac{\pi }{3} $