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If a=\(\frac { 1 }{ 4 } +i\frac { \sqrt { 3 } }{ 4 } \) and \(z=x+iy\), then \(\sin ^{ -1 }{ { \left| z \right| }^{ 2 } } +\cos ^{ -1 }{ (a\bar { z } } +\bar { a } z-2)\) equals to,

\((A)0\)

\((B)\frac { \pi }{ 4 } \)

\((C)\frac { \pi }{ 2 } \)

\((D)\frac { 3\pi }{ 2 } \)

Note by Anandhu Raj
2 years, 4 months ago

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We can write a and z as follows \[a=\frac {1}{2}e^{\frac {i \pi}{3}}\] \[z=|z|e^{i\theta }\]

\[a \overline{z}+z \overline{a}-2 =\frac {|z|}{2} (e^{(-i )\frac {\pi}{3} }e^{i \theta} + e^{(i \frac {\pi}{3}} e^{(-i )\theta})-2 \]

\[=|z| \cos (\theta-\frac {\pi}{3})-2\]

Now using domain of the inverse functions

\[-1\leq |z|^{2} \leq 1\] \[-1 \leq |z| \cos (\theta-\frac {\pi}{3})-2 \leq 1\]

From these it is easy to see that \[|z|= 1 and \cos (\theta-\frac {\pi}{3})=1\]

So we get the answer as \[sin^{-1}(1)+cos^{-1}(-1)= 3\pi /2 \]which is option D. Enjoy.

Satvik Choudhary - 2 years, 4 months ago

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Thank you :)

Anandhu Raj - 2 years, 4 months ago

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