Пусть z=a+bi;
![z^2=a^2-b^2+2abi z^2=a^2-b^2+2abi](https://tex.z-dn.net/?f=z%5E2%3Da%5E2-b%5E2%2B2abi)
0, \\ b^2=4;\\ b=+-2;\\ a=-+1;\\ \\ z=-1+2i;z=1-2i" alt="a^2-b^2+2abi+3+4i=0;\\ a^2-b^2+3=0\\ 2abi+4i=0\\ ab=-2\\ a=-2/b\\ 4/b^2-b^2+3=0;\\ 4-b^4+3b^2=0\\ b^4-3b^2-4=0;\\ b^2=(3+-\sqrt{3^2+4*4})/2=(3+-5)/2;\\ b^2>0, \\ b^2=4;\\ b=+-2;\\ a=-+1;\\ \\ z=-1+2i;z=1-2i" align="absmiddle" class="latex-formula">
представим
в тригонометрическом виде:
;
![z=r*(cos\frac{-\pi}{6}+isin\frac{-\pi}{6}); z=r*(cos\frac{-\pi}{6}+isin\frac{-\pi}{6});](https://tex.z-dn.net/?f=z%3Dr%2A%28cos%5Cfrac%7B-%5Cpi%7D%7B6%7D%2Bisin%5Cfrac%7B-%5Cpi%7D%7B6%7D%29%3B)
По формуле Муавра,
![z^4=r^4*(cos(4*\frac{-\pi}{6})+isin(4*\frac{-\pi}{6}))=1(-0,5-i*\sqrt3/2)=-1/2-i\sqrt3/2 z^4=r^4*(cos(4*\frac{-\pi}{6})+isin(4*\frac{-\pi}{6}))=1(-0,5-i*\sqrt3/2)=-1/2-i\sqrt3/2](https://tex.z-dn.net/?f=z%5E4%3Dr%5E4%2A%28cos%284%2A%5Cfrac%7B-%5Cpi%7D%7B6%7D%29%2Bisin%284%2A%5Cfrac%7B-%5Cpi%7D%7B6%7D%29%29%3D1%28-0%2C5-i%2A%5Csqrt3%2F2%29%3D-1%2F2-i%5Csqrt3%2F2)