Given x ∈∈ R and x << 0, therefore the number is −x+0i and this will be a point inthe second quadrant as x is negative
We have arg(−x+0i=θ= tan−1∣∣yx∣∣=tan−1∣∣0−1∣∣=tan−10=0
We have in the second quadrant the principal value of argument= Π−θ=Π−0=Π
Hence Arg. ( x ) , x ∈ R and x < 0 is π