I am confused on when to perform calculations on complex numbers using the argument or principal argument of a complex number.

I know the following properties do not necessarily hold for the principal argument:



I was asked to show $\frac{\pi}{4}=4\tan^{-1}(\frac{1}{5})-\tan^{-1}(\frac{1}{239})$

Using $(1+i)(5-i)^4$

First I found the argument, $\arg(1+i)(5-i)^4$

which was $\arg(1+i)(5-i)^4=\{-\tan^{-1}(\frac{1}{239})+2k\pi:k \in \mathbb{Z}\}$

Then I found using the additive property of arg,

$\text{arg}(1+i)=\{\frac{\pi}{4}+2k\pi:k\in \mathbb{Z}\}$

and $\text{arg}(5-i)^4=\{-4 \tan^{-1}(\frac{1}{5})+2k\pi:k \in \mathbb{Z} \} $

I then avoided the additive terms $2k\pi$ and concluded this implies


My professor said we were supposed to use the principal argument to perform the calculations. I did not because $\text{Arg}z_1z_2=\text{Arg}z_1+\text{Arg}z_2$ is not necessarily true.

Because of this I am very confused and do not know when I should and should not use the principal argument in calculations.Also my professor never went over these properties of the principal argument. I noticed also the book likes to use $\text{arg}$ without taking on values that are periodic with $2\pi$

I asked the professor,"can we ever abuse notation and let $\text{arg}$ take on a single value." He said no. This makes me more confused, if we are asked to perform a calculation like the one above where $\text{Arg}z_1z_2=\text{Arg}z_1+\text{Arg}z_2$ is not satisfied.


Using arg one can show a congruence relation, i.e. $A \equiv B \mod 2 \pi$ you can convert this to an equality if you provide a bounds checking step, which must be done. You a right to be concerned about using Arg. Incidently you shouldn't recycle the variable k for different equations. It won't necessarily cancel.


Note that $arg(z) =Arg(z) + 2n \pi$, $n \in \mathbb{Z}$. In other words, $arg(z) =Arg(z) mod 2 \pi$.

Hence, $Arg(z_1 z_2) = (Arg(z_1) + Arg(z_2)) mod 2 \pi$.

Similarly, $Arg(z^n) = (n Arg(z) )mod 2 \pi$.

You already considered $(1+i)(5-i)^4$. Using above relations,

$$Arg[(1+i)(5-i)^4] =(Arg(1+i)+ 4. Arg(5-i)) mod 2 \pi $$

$$Arg(956-4i) = (arctan(1) + 4 arctan(-\frac{1}{5}))mod 2 \pi$$

$$arctan(- \frac{1}{239})= (\frac{\pi}{4} + 4. arctan(-\frac{1}{5}))mod 2 \pi$$

So, $$arctan(- \frac{1}{239}) - 4. arctan(-\frac{1}{5}) + 2\pi k = \frac{\pi}{4},$$ where $k \in \mathbb{Z}.$

Note that , $arctan(\frac{1}{239}) = 2 \pi k' - arctan(- \frac{1}{239})$.

Similarly, $4.arctan(\frac{1}{5}) = 2 \pi k_0 - 4.arctan(- \frac{1}{5})$, $|k_0|\geq 4$.

This gives,$$ 4. arctan(\frac{1}{5}) -arctan(\frac{1}{239}) = \frac{\pi}{4} $$.

  • $\begingroup$ so was the professor wrong to have said, just calculate the principle argument of each component and add the results? $\endgroup$ – user736276 Jan 24 '20 at 1:32
  • $\begingroup$ @68e1515 You can't certainly add principle arguments as you would do with $arg(z)$. You can add 'em in $mod 2 \pi$ . $\endgroup$ – SL_MathGuy Jan 24 '20 at 1:36
  • $\begingroup$ He said in this particular problem $\text{Arg}z_1z_2=\text{Arg}z_1+\text{Arg}z_2$ Is this true or do we still have to add them mod$2\pi$? $\endgroup$ – user736276 Jan 24 '20 at 1:38
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    $\begingroup$ @68e1515 en.wikipedia.org/wiki/Argument_(complex_analysis) Sorry for the confusion. I meant $Arg (z) = arg (z)$ iff $x,y >0$ ($z =x + iy$) $\endgroup$ – SL_MathGuy Jan 24 '20 at 2:03
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    $\begingroup$ I found an answer to this problem. I have a pdf of brown and Churchill complex variables and it states $Argz_1z_2=Argz_1+Argz_2$ if Re$z_1>0$ and Re$z_2>0$ so the professor was right $\endgroup$ – user736276 Jan 24 '20 at 2:26

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