MHT CET · Maths · Complex Number
The argument of \(\frac{1+i \sqrt{3}}{\sqrt{3}+i}, i=\sqrt{-1}\) is
- A \(\frac{\pi}{3}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{6}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{6}\)
Step-by-step Solution
Detailed explanation
\(\text {Let } z=\frac{1+i \sqrt{3}}{\sqrt{3}+i} \)
\( = \frac{(1+i \sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)} \)
\( \therefore z =\frac{\sqrt{3}}{2}+\frac{1}{2} i\)
Argument of
\(z =\tan ^{-1}\left(\frac{b}{a}\right) \)
\(=\tan ^{-1}\left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \)
\(=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
\(=\frac{\pi}{6}\)
\( = \frac{(1+i \sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)} \)
\( \therefore z =\frac{\sqrt{3}}{2}+\frac{1}{2} i\)
Argument of
\(z =\tan ^{-1}\left(\frac{b}{a}\right) \)
\(=\tan ^{-1}\left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \)
\(=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
\(=\frac{\pi}{6}\)
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