Find the argument of $\displaystyle -\frac 12+iT$.

  • Here the point $\displaystyle -\frac 12+iT$ lies in the 3rd quadrant. So, $\displaystyle \arg \left(-\frac 12+iT\right)=\pi +\arctan(-2T).$
  • Again,

    $\displaystyle \arg \left(-\frac 12+iT\right)=\arg \left(i(T+i/2)\right)=\arg (i)+\arg (T+i/2)=\pi/2+\arctan(1/2T)$.

Which one is correct?

I know that $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$ holds only for general values, not necessarily for pricipal argument.

But in my book it is given that, $\displaystyle \arg \left(-\frac 12+iT\right)=\pi/2+\arctan(1/2T)$. That's why I'm confused !!


Both are obviously correct up to multiples of $\pi$. Now because of $T>0$ the first one, $\pi-\arctan(2T)$, has a value in $[\frac\pi2,\pi]$ and the second one, $\frac\pi2+\arctan(\frac1{2T})$, has its value also in $[\frac\pi2,\pi]$, so both formulas give identical results.

  • $\begingroup$ But how can I write $\arg (z_1z_2)=\arg(z_1)+\arg(z_2) ?$ $\endgroup$ – Empty Nov 7 '18 at 16:38
  • 1
    $\begingroup$ You can not, as you observed. But $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)+2k\pi$ for one of $k\in\{-1,0,1\}$ and it is usually easy to argue which $k$ to take. Using the $\arctan$ instead of the argument (=atan2) reduces the period to $\pi$, but still the possible values are far enough away from each other that finding the correct one is no unsurmountable obstacle. $\endgroup$ – LutzL Nov 7 '18 at 16:43

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