Problems love to argue with people

Algebra Level 4

$\large \text{Arg} (\bar{z}) =- \text{Arg}(z)$

Does the above property hold for each complex number $z$ ?

Clarifications:

$\text{Arg}(z)$ is the principal argument of complex number $z$ , with a range of $(-\pi, \pi ]$ .
$\bar z$ is the complex conjugate of complex number $z$ .

No Yes

1 solution

Rishabh Deep Singh
Mar 19, 2016

Argument of a complex number is $-\pi <Arg(Z)\le \pi$

Now Consider a complex number $\overline { Z } =-1$ having $Arg(\overline { Z } )=\pi$ and $Re\left( \overline { Z } \right) =-1\quad and\quad Im\left( \overline { Z } \right) =0$

And also $Z=-1$ .

Therefore from given equation $Arg\left( \overline { Z } \right) =\quad -\quad Arg\left( { Z } \right)$

$Arg\left( { Z } \right) =\quad -\quad Arg\left( \overline { Z } \right)$

$Arg\left( { Z } \right) =\quad -\left( \pi \right) =-\pi$

But we know that Argument of a complex number is $-\pi <Arg(Z)\le \pi$

Therefore it is not correct for all complex numbers.

Moderator note:

There are times when it is better to think of a function as multi-valued, which is what allows us to say that $Arg(1) = 0, 2 \pi, 4 \pi, \ldots$ , and thus satisfy a more general statement.

The principal Argument can also be defined Theta in [0, 2Pi) this proof fails for that definition since 0 is unsigned and -Pi is in the range. I find this problem very strange since it relies on the definition of the bounds of the principal argument to determine the method you disprove the statement with. Moreover the bounds I gave work for the end points, but fail for all the interior, the bounds you gave work for all interior but fail for -Pi and collectively the cover the whole circle. Which means that for any Arg(z) = -Arg(conj(z)) I can find a definition of Arg(z) such that the equation is true, but if I have one definition it fails.

Russell Hart - 5 years, 2 months ago

Agreed. It is just a matter of technicality

Agnishom Chattopadhyay - 5 years, 2 months ago

Thanks. I've edited the problem to indicate the actual range of the "principal argument".

Yes, there are times when it is better to think of a function as multi-valued, which is what allows us to say that $Arg(1) = 0, 2 \pi, 4 \pi, \ldots$ , and thus satisfy a more general statement.

Calvin Lin Staff - 5 years, 2 months ago

You should change the last line to "not necessarily correct for all complex numbers"

Agnishom Chattopadhyay - 5 years, 2 months ago

Truly quirky and without any practical relevance!

Andreas Wendler - 5 years, 2 months ago

Problems love to argue with people

1 solution

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