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

Mathematics Asked by cocacola on December 16, 2020

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 = $$frac{pi}{3}$$

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, $$z=(2text{cis}(-dfrac{pi}{3}))^6=2^6text{cis}(-2pi)=2^6text{cis}(boldsymbol{0})=2^6=64$$

Correct answer by talbi on December 16, 2020

$$Re^{itheta} =(1-isqrt{3})^6$$

$$Rightarrow Re^{itheta}=2^6(frac{1} {2} - ifrac{sqrt{3}}{2})^6$$

$$Rightarrow Re^{itheta}= 2^6 (e^{frac{-ipi}{3}}) ^{6}$$

$$Rightarrow Re^{itheta}= 2^6e^{-i2pi}$$

$$Rightarrow R=2^6$$, $$theta=-2pi+2kpi$$

$$K=1$$

So:

$$R=64$$ and $$theta=0$$

Answered by Anas chaabi on December 16, 2020

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.)

Answered by user247327 on December 16, 2020