Real and imaginary part of $ (1-i\sqrt{3})^6$

Question

i am a bit stuck here.

As the title says i try to find out how to write complex numbers like for example$$ (1-i\sqrt{3})^6$$ in the normal representation$$ z = x + i*y$$ I already found out that the polar representation of complex numbers will come in handy here, but i can't make the conclusion at the moment.

How can i get from here to the polar representation? How do i get the real and imaginary part from the polar representation? If you have a hint, can you please just leave a quick post here, thanks.

Answer 1

The modulus of $1-i\sqrt{3}$ is $\sqrt{1+3}=2$, so you can write $$ 1-i\sqrt{3}=2\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right) = 2(\cos(-\pi/3)+i\sin(-\pi/3)) $$ Can you compute the sixth power, now?

Answer 2

Hints:

$$1-i\sqrt3=2e^{-\frac{\pi i}3}\implies (1-i\sqrt3)^6=\left(2e^{-\frac{\pi i}3}\right)^6=\ldots$$ Another forceful way:

$$(1-i\sqrt3)(1-i\sqrt3)=-2-2\sqrt3\,i=-2(1+\sqrt3\,i)\implies $$

$$(1-\sqrt3\,i)^3=-2(1+\sqrt3\,i)(1-\sqrt3\,i)=-2\cdot4=-8\implies$$

$$(1-\sqrt3\,i)^6=\left((1-\sqrt3\,i)^3\right)^2=\ldots$$

Answer 3

we know $$w=\dfrac{-1+\sqrt{3}i}{2}\Longrightarrow w^2+w+1=0\Longrightarrow w^3=1$$ so $$(1-i\sqrt{3})^6=(\sqrt{3}i-1)^6=2^6\cdot\left(\dfrac{\sqrt{3}i-1}{2}\right)^6=2^6w^6=2^6\cdot 1^2=64$$