Question 238652
First you have to convert *[Tex \LARGE z=x+yi] into polar form *[Tex \LARGE z=r\left(\cos(\theta)+i\sin(\theta)\right)]. 


To do that, use the formulas *[Tex \LARGE r=\sqrt{x^2+y^2}] and *[Tex \LARGE \theta=\tan^{-1}\left(\frac{y}{x}\right)]



In this case, we're given *[Tex \LARGE z=1+i] which means that {{{x=1}}} and {{{y=1}}}. So *[Tex \LARGE r=\sqrt{1^2+1^2}=\sqrt{2}] and *[Tex \LARGE \theta=\tan^{-1}\left(\frac{1}{1}\right)=\tan^{-1}\left(1\right)=\frac{\pi}{4}]


So this means that the rectangular expression {{{z=1+i}}} is equivalent to the polar form *[Tex \LARGE z=\sqrt{2}\left(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\right)].


From here, we can now use De Moivre's Theorem. De Moivre's Theorem states that if *[Tex \LARGE z=r\left(\cos(\theta)+i\sin(\theta)\right)], then *[Tex \LARGE z^n=r^n\left(\cos(n\theta)+i\sin(n\theta)\right)]


Since *[Tex \LARGE z=\sqrt{2}\left(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\right)], using De Moivre's Theorem gets us




*[Tex \LARGE z^{12}=\left(\sqrt{2}\right)^{12}\left(\cos(12*\frac{\pi}{4})+i\sin(12*\frac{\pi}{4})\right)]



*[Tex \LARGE z^{12}=2^{6}\left(\cos(3\pi)+i\sin(3\pi)\right)] ... Evaluate {{{sqrt(2)}}} to the 12th power to get {{{(sqrt(2))^12=(2)^((1/2)(12))=2^6}}}



*[Tex \LARGE z^{12}=64\left(\cos(3\pi)+i\sin(3\pi)\right)] ... Evaluate 2 to the 6th power to get 64



*[Tex \LARGE z^{12}=64\left(-1+i*0\right)] ... Evaluate the trig functions.



*[Tex \LARGE z^{12}=64\left(-1\right)] ... Simplify.



*[Tex \LARGE z^{12}=-64] ... Multiply



Because we let {{{z=1+i}}}, this means that {{{(1+i)^12=-64}}}