Find the argument of $-1/2+iT$

Question

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 !!

Answer 1

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.