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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,

$(A)0$

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

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

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

#ComplexNumbers #InverseTrigonometricFunctions

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 - 5 years, 10 months ago

Thank you :)

Anandhu Raj - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$