Raising complex number to high power - Cartesian form My question is about raising a complex number to a high power, I know how to do that with De Moivre law, but i need to get the result in cartesian form, like $z=x+iy$. and without trigonometric terms.
The problem exactly is:

Write the following complex number in the following form $z=x+iy$:
  $$(3-2i)^3\cdot(1-i)^9$$

I searched alot the web to find an understandible explanation to solve the problem, but still i have no idea how to do that.
Is there a special formula to raise a complex number to high power?
Thanks for help!!
 A: For the factor of $(1 - i)^9,$ I think the de Moivre form yields some insight, because $1 - i = \sqrt2 e^{-i\pi/4}.$
Hence 
$$(1 - i)^9 = 2^{9/2} e^{-i9\pi/4} = 16\sqrt2 e^{-i\pi/4} = 16 - 16i.$$
For the factor of $(3-2i)^3$ I am not convinced by the other answers that
computing $\cos(3\arctan(-2/3))$ is simpler than just doing two complex multiplications by the algebraic method.
A: If we want proceed in cartesian form we need to expand
$$(3-2i)^3\cdot(1-i)^9=(3-2i)\cdot (3-2i)\cdot(3-2i)\cdot (1-i)\cdot \ldots \cdot (1-i)$$
The best way is by exponential form
$$z=x+iy =re^{i\theta}\implies z^n=r^ne^{in\theta}$$
with


*

*$r=\sqrt{x^2+y^2}$

*$\theta = \arctan (y/x)\, [+\pi]$ (for $x\neq 0$)

A: Hint: $(1-i)^2=-2i$ then $(1-i)^8=16$. then
$$(3-2i)^3(1-i)^9=16\color{blue}{(3-2i)^2}\color{red}{(3-2i)(1-i)}$$
A: You can use the binomial formula and exploit simplifications.
$$(3-2i)^3=3^3-3\cdot3^2\cdot2\,i-3\cdot3\cdot2^2+2^3\,i=-9-46i,$$
$$(1-i)^9=(1-i)(1-i)^8=(1-i)(-2i)^4=16(1-i).$$
But in general, there is no shortcut.

Algorithmically speaking, the most efficient way to raise to a high power is by successive squarings, which is better than repeated multiples, but for moderate exponents, you won't see a big difference.

I don't think that the trigonometric form is of much  use because
$$(\sqrt{13})^3(\sqrt2)^9e^{-i(3\arctan2/3+9\pi/4)}$$
doesn't tell you much more than $(3-2i)^3(1-i)^9$.
A: If you use DeMoivre's Formula and get $r(\cos x + i\sin x)$ then simply distribute the $r$ to get $$r\cos x + (r\sin x)i$$
Now just evaluate the cosine and sine function and you're left with a complex number in $a+bi$ form.
