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

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.
• But how can I write $\arg (z_1z_2)=\arg(z_1)+\arg(z_2) ?$ – Empty Nov 7 '18 at 16:38
• 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. – LutzL Nov 7 '18 at 16:43