Question

If $- \pi < \arg \left( z \right) < \- \dfrac{\pi }{2}$ then $\arg \left( {\bar z} \right) - \arg \left( {\overline { - z} } \right) =$
(1) $\pi$
(2) $- \pi$
(3) $\dfrac{\pi }{2}$
(4) $\dfrac{{ - \pi }}{2}$

Answer 1

First we write the complex number $z$ in its polar form. After that we will find all the terms that are required in the question i.e., $- z$ , conjugate of $z$ and $- z$ (represented as $\bar z$ and $\overline { - z}$ ) and argument of $\bar z$ and $\overline { - z}$ , represented by $\arg \left( {\bar z} \right)$ and $\arg \left( {\overline { - z} } \right)$ . Then substitute the value in the given condition to get the result.
Properties used:
(i) If $z = a + ib$ ,then $- z = - a - ib$
(ii) If $z = a + ib$ ,then $\bar z = a - ib$
(iii) $\arg \left( z \right) = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$

Complete answer:
We have to find $\arg \left( {\bar z} \right) - \arg \left( {\overline { - z} } \right){\text{ }} - - - \left( A \right)$
Let $z$ be a complex number.
So, in polar form, $z = \cos \theta + i\sin \theta {\text{ }} - - - \left( 1 \right)$
And we know that,
if $z = a + ib$ ,then $- z = - a - ib$
From equation $\left( 1 \right)$
$- z = - \cos \theta - i\sin \theta {\text{ }} - - - \left( 2 \right)$
Now, we will find the conjugate of $z$ and $- z$
i.e., If $z = a + ib$ ,then $\bar z = a - ib$ , means we have to change the sign of coefficient of iota.
So, from equation $\left( 1 \right)$
$\bar z = \cos \theta - i\sin \theta$
And we know that, $\sin \left( { - \theta } \right) = - \sin \theta$ and $\cos \left( { - \theta } \right) = \cos \theta$
$\therefore \bar z = \cos \left( { - \theta } \right) + i\sin \left( { - \theta } \right){\text{ }} - - - \left( 3 \right)$
And from equation $\left( 2 \right)$
$\overline { - z} = - \cos \theta + i\sin \theta$
As we can see that $\cos$ is negative and $\sin$ is positive, it is possible in 2nd quadrant
So, it can also be written as,
$\overline { - z} = \cos \left( {\pi - \theta } \right) + i\sin \left( {\pi - \theta } \right){\text{ }} - - - \left( 4 \right)$
Now, we will find argument of $\bar z$ and $\overline { - z}$
we know that $\arg \left( z \right) = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$
$\therefore$ from $\left( 3 \right)$ , $\arg \left( {\bar z} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\sin \left( { - \theta } \right)}}{{\cos \left( { - \theta } \right)}}} \right)$
$\Rightarrow \arg \left( {\bar z} \right) = {\tan ^{ - 1}}\left( {\tan \left( { - \theta } \right)} \right)$
And we know that,
${\tan ^{ - 1}}\left( {\tan \left( x \right)} \right) = x$
$\Rightarrow \arg \left( {\bar z} \right) = - \theta$
Now from $\left( 4 \right)$ , $\arg \left( {\overline { - z} } \right) = {\tan ^{ - 1}}\left( {\dfrac{{\sin \left( {\pi - \theta } \right)}}{{\cos \left( {\pi - \theta } \right)}}} \right)$
$\Rightarrow \arg \left( {\overline { - z} } \right) = {\tan ^{ - 1}}\left( {\tan \left( {\pi - \theta } \right)} \right)$
$\Rightarrow \arg \left( {\overline { - z} } \right) = \pi - \theta {\text{ }}$
Now, substitute the value of $\arg \left( {\bar z} \right)$ and $\arg \left( {\overline { - z} } \right)$ in $\left( A \right)$
$\Rightarrow \arg \left( {\bar z} \right) - \arg \left( {\overline { - z} } \right) = - \theta - \left( {\pi - \theta } \right)$
$\Rightarrow \arg \left( {\bar z} \right) - \arg \left( {\overline { - z} } \right) = - \pi$
Hence option (2) is correct.

Note:
To solve this question, first we need to know about the polar form of complex numbers. We should also know about the basic properties of conjugate and argument of complex numbers. Also, while solving the question in which condition is given for angles, we should take care of that condition in the complete solution. Like in this question it is given that $- \pi < \arg \left( z \right) < \- \dfrac{\pi }{2}$ $\arg \left( z \right) = {\tan ^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)$
$\Rightarrow \arg \left( z \right) = {\tan ^{ - 1}}\left( {\tan \left( \theta \right)} \right) = \theta$
It means that, in the given question
$- \pi < \theta < \- \dfrac{\pi }{2}$
$\Rightarrow \pi < \- \theta < \dfrac{\pi }{2}$ which satisfies the condition for cos and sin in equation $\left( 3 \right)$
Also, $0 < \pi - \theta < \dfrac{\pi }{2}$ which satisfies the condition for cos and sin in equation $\left( 4 \right)$