Confused about how to find the value of $(1+i)^{12}$

Question

I am learning de Moivre’s theorem, which states that

(cos $\theta +i$ sin $\theta)^n =$ cos $n\theta + i$ sin $n\theta$

In my book, there is this question: $(1+i)^{12}$

And the book provides the following solution:

$(1+i)^{12}$

$=(\sqrt{2})^{12}\big(\displaystyle\frac{1}{\sqrt2}+i \frac{1}{\sqrt2}\big)^{12}$

$=64\big($ cos $\displaystyle\frac{\pi}{4} \color {red}-i$ sin $\displaystyle\frac{\pi}{4}\big)^{12}$

$=64 ($cos $3\pi+ i$ sin $3\pi\big)$

$=64 ($cos $\pi+i$ sin $\pi)$

$=-64$

I do not understand the second line of the solution: why is the $\color{red}{red}$ sign negative, and not positive?

Answer 1

The red negative sign is a typo. It should be positive and the working will be correct.

Answer 2

It was typo, but the red minus gives the same result: $$64\left(\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\right)^{12}=64\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)^{12}=$$ $$=64(\cos(-3\pi)+i\sin(-3\pi))=-64.$$

I think the following is a bit of better. $$(1+i)^{12}=((1+i)^2)^6=(1+2i-1)^6=(2i)^6=-64.$$