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

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3 Answers 3

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 $-\frac{\pi}{2}.$

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

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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 $n=0$)

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