Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $z = (4\sqrt{3} - 4 i)^3$, determine $\arg z$. How to find out this $\arg z$? i need help. thanks a lot...

share|cite|improve this question

We can also use DeMoivre's Theorem here: we have $$(4\sqrt{4}-4i)^3=4^3\left(2(\frac{\sqrt{3}}{2}-\frac{1}{2}i)\right)^3=8^3(\cos(-30^\circ)+i\sin(-30^\circ))^3.$$ Now, by DeMoivre's Theorem, this reduces to $$8^3(\cos(-90^\circ)+i\sin(-90^\circ)),$$ so the principal argument is $-90^\circ$ or $-\frac{\pi}{2}.$

share|cite|improve this answer

Hint: $\arg (z_1z_2)=\arg (z_1)+\arg (z_2)$

share|cite|improve this answer
This is not always true for the principal argument. – lab bhattacharjee Dec 15 '12 at 15:56
Once you know any value of $\arg(z)$, it's easy to find $\mathop{\mathrm{Arg}}(z)$. – Hurkyl Dec 15 '12 at 15:57
I think principal argument is usually denoted by 'Arg' – pritam Dec 15 '12 at 15:57

Lets use $Arg(z),arg(z)$ as the principal & the general argument $z$ respectively.

$arg(4\sqrt3-4)$ is $n\pi+\arctan\left( \frac{-4}{4\sqrt3}\right)=n\pi-\arctan\frac1{\sqrt3}=n\pi-\frac{\pi}6$ where $n$ is any integer.

$arg\left((4\sqrt3-4)^3\right)$ is $3(n\pi-\frac{\pi}6)=3n\pi-\frac{\pi}2$ as $arg(z\cdot w)=arg(z)+arg(w)$

As the principal argument in $(-\pi,\pi), Arg(4\sqrt3-4)^3$ will be $-\frac{\pi}2$ (putting $n=0$)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.