Evaluate $(-\sqrt3 + i)^9$

Question

Evaluate $(-\sqrt3 + i)^9$

So I used the binomial theorem, but it took quite a while to evaluate all those out. However, seeing as how this is a problem on a competition, I was wondering if there was a faster way to do this.

Answer 1

Hint :

$$-\sqrt 3+i =2\left( \frac{-\sqrt 3}{2} +\frac 12 i\right)=2 \left( \cos \left( \frac{5\pi}{6} \right) +i\sin \left( \frac{5\pi}6 \right) \right) = 2e^{i5\pi /6}$$

Answer 2

Hint: Write the number in polar form. Evaluate the norm and the tangent of angle in complex plane. Then use the fact that the angle is multiplied $n$ times when raised to power $n$.

Answer 3

If you know your way around complex numbers and De Moivre's theorem, you easily get that the answer is equal to -512i. Hint: rewrite it as (2 ((cos(5π/6) + sin (5π/6))^9 and then apply the Theorem.

Answer 4

HINT

Use trigonometric or polar form such that

$$z=r e^{i\theta}\implies z^n= r^ne^{in\theta}$$

Answer 5

Hints :- Take i common and you get the well known $2i\omega$ and we know that cube of $\omega$ is 1 and $i^9=i $. I hope you can use them to get the final answer.

Answer 6

Following up on OP's approach, without using the polar form:

So I used the binomial theorem, but it took quite a while to evaluate all those out

It's enough to calculate the first couple of powers, which goes rather quickly.

Let $\,z=-\sqrt3 + i\,$, then $\,z^2=3 - 2 \sqrt{3}i+i^2=2(1-i\sqrt{3})=2i(-\sqrt{3}-i)=2i \bar z$. It follows that $\,z^3=z \cdot 2i \bar z = 2i |z|^2=8i\,$, and $\,z^9=\big(z^3\big)^3 = (8i)^3=\ldots\,$