Question

If \[ - \pi < \arg \left( z \right) < - \dfrac{\pi }{2}\], then \[\arg \left( {\bar z} \right) + \arg \left( { - \bar z} \right)\] is equal to
A. \[\pi \]
B. \[ - \pi \]
C. \[\dfrac{\pi }{2}\]
D. \[ - \dfrac{\pi }{2}\]

Answer 1

Hint: Let \[z = x + iy\]. Argument of the complex number \[z\] is given. Using that identify the position of \[z\]. Then find its conjugate \[\bar z\]and negative of \[\bar z\]. After that find their arguments and add them to find the required answer.



Formula Used:If \[z = x + iy\], then \[\bar z = x - iy\]
The point corresponding to a complex number lies in first quadrant if \[0 < \arg \left( z \right) < \dfrac{\pi }{2}\], lies in second quadrant if \[\dfrac{\pi }{2} < \arg \left( z \right) < \pi \], lies in third quadrant if \[ - \pi < \arg \left( z \right) < - \dfrac{\pi }{2}\], lies in fourth quadrant if \[ - \dfrac{\pi }{2} < \arg \left( z \right) < 0\].



Complete step by step solution:Let \[z = x + iy\], where \[x\] and \[y\] are real numbers and \[i = \sqrt { - 1} \]
Given that \[ - \pi < \arg \left( z \right) < - \dfrac{\pi }{2}\]
It means \[z\] lies in third quadrant i.e. the values of both \[x\] and \[y\] are negative.
Let \[\arg \left( z \right) = - \pi + \theta \], where \[0 < \theta < \dfrac{\pi }{2}\]
Now, the conjugate of \[z\] is \[\bar z = x - iy\], which lies in second quadrant.
So, \[\arg \left( {\bar z} \right) = \pi - \theta \]
Again, \[ - \bar z = - \left( {x - iy} \right) = - x + iy\], which lies in first quadrant.
So, \[\arg \left( { - \bar z} \right) = \theta \]
Now, \[\arg \left( {\bar z} \right) + \arg \left( { - \bar z} \right) = \left( {\pi - \theta } \right) + \left( \theta \right) = \pi - \theta + \theta = \pi \]



Option ‘A’ is correct



Note: Argument of a complex number is the angle between the real axis and the line passes through origin and the point corresponding to the complex number. Many students can’t identify actual location of a complex number due to they can’t remember the ranges of the argument of the complex number quadrant-wise.