How to express $(-1+ \sqrt{3}i)^5$ in rectangle form? The question is "Express $(-1+ \sqrt{3}i)^5$ in rectangle form."
I've never heard of rectangle form before and it's apparently gonna be on the final exam. I couldn't find anything on google to help.
 A: \begin{align*}
  (-1+\sqrt{3}\,i)^{5} &=
  2^{5} \left( -\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{5} \\
  &= 2^{5} (e^{2\pi i/3})^{5} \\
  &= 2^{5} e^{10\pi i/3} \\
  &= 32\, e^{4\pi i/3} \\
  &= 32\left( -\frac{1}{2}-\frac{\sqrt{3}}{2} i\right) \\
  &= -16-16\sqrt{3} \, i
\end{align*}
A: Rectangular form is simply the standard $a+ib$ form for the number (as opposed to another form, say polar). Therefore you simply need to multiply out to get $-\sqrt{3i}^5 = 11.0227 + 11.0227i$
A: "rectangular form" is probably the form you are most familiar with.  $z = x + y\,i $
An alternative would be  "polar form." $z = \rho(\cos\theta + i\sin\theta)$
Now one of the nice things about polar form is "DeMoivre's theorem"
$z^n = \rho^n(\cos n\theta + i\sin n\theta)$
Convert $(-1 + i\sqrt{3})$ into polar form.
Apply DeMoivre's theorem.
Convert from polar form to rectangular form.
By the way, What Ng Chung Tak has done above is completely correct.  However, if you don't know "rectangle form", I highly doubt that you know Euler's identity or exponential forms of complex numbers.
