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# Help Needed!

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 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. · 1 year, 12 months ago