Finding the value of $(-1+i)^{10}$ by using the trigonometric form's expression I am trying to find out what is the value of $(-1+i)^{10}$ by using the trigonometric expression, which should simplify it by avoiding to multiply in binomial form. So far I got this (sin==sen, and excuse my shoddy writing):

But every other calculator I used to test the result says that it's -$32i$, and not $32i$, which is driving me mad! I can only guess that I messed u on the argument($\alpha$)
Any help is welcome!
Edit: Solved, I hadn't added the fractions properly, see comment chain below post
 A: $$-1+i=\sqrt2\exp(3\pi i/4)$$
so
$$(-1+i)^{10}=32\exp(15\pi i/2)=-32i.$$
Alas,
$$\sqrt2\exp(5\pi i/4)=-1-i.$$
A: Write $z=-1 + i$ in polar form. That is $z=\sqrt{2}e^{i\frac{3}{4}\pi}$. Then, $z^{10} = 2^5 e^{i\frac{15}{2}\pi} $. But that an angle of $\frac{15}{2}\pi$ is the same as $\frac{3}{2} \pi$, and $e^{i\frac{3}{2}\pi} = \cos(\frac{3}{2}\pi) + i\sin(\frac{3}{2}\pi) = -i$. 
Then, $z^{10} = -32i$
A: $$N=(-1+i)^{10}=(\sqrt{2} e^{3i\pi/4})^{10}=32 e^{15i\pi/2}=32~ e^{7i\pi} ~e^{\pi/2}= -32i.$$
A: Cartesian to polar conversion in complex plane
$$a+ib =\sqrt{a^2+b^2} e^{i\cdot \tan^{-1}(b/a)}$$
Raising to $n^{th}$power 
$$(a+ib)^n =(a^2+b^2)^{n/2} e^{i\cdot n\cdot \tan^{-1}(b/a)}$$
$$(-1+i)^{10}=(\sqrt{2} e^{i \cdot 3 \pi/4})^{10}=32 e^{15i\pi/2}=32~ e^{12 i\pi/2} \,e^{3\pi/2}=32 (1) e^{3\pi/2} =-32i$$
as the radius vector after rotation lands on the negative y-axis in the complex plane.
A: $(-1+i)$ is a vector of length $\sqrt{2}$ and an angle of $135$ degrees. The resulting vector has a length of $\sqrt{2}^{10} = 2^5 = 32$ and an angle of $10 \cdot135 = 1350 \equiv 270 \pmod{360}$ degrees, so yes, it's $-32i$.
A: Hint:
Use https://en.m.wikipedia.org/wiki/Atan2#Definition_and_computation
arg$(-1+i)=\arctan\dfrac1{-1}+\pi=\dfrac{3\pi}4$
arg$(-1+i)^{10}=10\cdot\dfrac{3\pi}4=\equiv-\dfrac\pi2\pmod{2\pi}$
