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If π<arg(z)<π2, then arg(z¯)+arg(z¯) is equal to
A. π
B. π
C. π2
D. π2


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



Formula Used:If z=x+iy, then z¯=xiy
The point corresponding to a complex number lies in first quadrant if 0<arg(z)<π2, lies in second quadrant if π2<arg(z)<π, lies in third quadrant if π<arg(z)<π2, lies in fourth quadrant if π2<arg(z)<0.



Complete step by step solution:Let z=x+iy, where x and y are real numbers and i=1
Given that π<arg(z)<π2
It means z lies in third quadrant i.e. the values of both x and y are negative.
Let arg(z)=π+θ, where 0<θ<π2
Now, the conjugate of z is z¯=xiy, which lies in second quadrant.
So, arg(z¯)=πθ
Again, z¯=(xiy)=x+iy, which lies in first quadrant.
So, arg(z¯)=θ
Now, arg(z¯)+arg(z¯)=(πθ)+(θ)=πθ+θ=π



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.