Evaluate $(\sqrt{3}-3i)^6$

Question

Evaluate $$(\sqrt{3}-3i)^6.$$

So I assume that we should write the following in polar form

$r=\sqrt{(\sqrt{3})^2+(-3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3},$

$\theta=\arctan{\frac{-3}{\sqrt{3}}}=-\frac{\pi}{3}.$

So $$(\sqrt{3}-3i)^6=[2\sqrt{3}e^{-i\frac{\pi}{3}+2\pi k}]^6=1728e^{-2\pi i+2\pi m},$$ where $m\in \mathbb{Z}.$

So $$1728e^{-2\pi i+2\pi m}=1728cos(-2\pi)=1728.$$

Is it correct?

There is no need to add $2 \pi k$ in the argument if you are evaluating integer powers of a complex number. Also, you missed $i$ in the exponent. — Math Lover, Aug 16 '17 at 17:33
The answer is correct, but your $2\pi k$ needs to be $2\pi k i$. — B. Goddard, Aug 16 '17 at 17:33

score 2 · Answer 1 · answered Aug 16 '17 at 17:35

2

Yes, it is correct. In fact $$(\sqrt{3}-3i)^6=3^3(1-\sqrt{3}i)^6=2^63^3(e^{-i\pi/3})^6=1728.$$ P.S. There is no need of the term $+2\pi k$.

answered Aug 16 '17 at 17:35

Robert Z

score 1 · Answer 2 · edited Aug 17 '17 at 11:21

1

You are right. It's $$27\cdot2^6\cos^6(-60^{\circ})=1728.$$

edited Aug 17 '17 at 11:21

InsideOut

answered Aug 16 '17 at 17:33

Michael Rozenberg

MAN-MADE · Answer 3 · 2017-08-16T17:45:37.690

0

$(\sqrt{3}-3i)^6=27(1-\sqrt{3}i)^6=27\times 2^6\Big(\dfrac{1}{2}-\dfrac{\sqrt3}{2}i\Big)^6=27\times64\space\omega^6=27\times 64$

$\omega$ is a non-real solution of the equation $x^6=1$

edited Aug 16 '17 at 17:45

answered Aug 16 '17 at 17:39

MAN-MADE

3 Answers3