Arg of $(1-isqrt{3})^6$. Did I do it right?

Question

Did I do it right?
$(1-isqrt{3})^6$
module = $sqrt{1+3}=2$
$(1-isqrt{3})^6=2^6cdot(frac{1}{2}-frac{sqrt{3}}{2}i)^6$ $implies$ Arg = $frac{pi}{3}$

talbi · Accepted Answer

I'm not sure what you did, but it is wrong as $(1-sqrt{3}cdot i)^6=boldsymbol{64}$ (Therefore the $text{Arg}$ is $0$).
To solve this, let $z=(1-sqrt{3}i)^6$. We'll turn $1-sqrt{3}cdot i$ into its polar form.
So, $|1-sqrt{3}cdot i|=sqrt{1^2+(sqrt{3})^2}=2$ like you've said. Also, $text{Arg}(1-sqrt{3}cdot i)=-dfrac{pi}{3}$
Finally, $z=(2text{cis}(-dfrac{pi}{3}))^6=2^6text{cis}(-2pi)=2^6text{cis}(boldsymbol{0})=2^6=64$

Anas chaabi · Answer

$Re^{itheta} =(1-isqrt{3})^6$
$Rightarrow $$Re^{itheta}=2^6(frac{1} {2} - ifrac{sqrt{3}}{2})^6$
$Rightarrow $$Re^{itheta}= 2^6 (e^{frac{-ipi}{3}}) ^{6}$
$Rightarrow $$ 
Re^{itheta}= 2^6e^{-i2pi} $
$Rightarrow$$ R=2^6$, $theta=-2pi+2kpi$
$K=1$
So:
$R=64 $ and $ theta=0$

user247327 · Answer

The argument of $1- isqrt{3}$ is $arctan(-sqrt{3})= -frac{pi}{3}$  So the argment of $(1-isqrt{3})^6$ is  $6times-frac{pi}{3}= -2pi$ which is equivalent to 0.
(Edited to correct negative sign, thanks to Ripi2.)

Arg of $(1-isqrt{3})^6$. Did I do it right?

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