Пусть z=a+bi;

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:
;

По формуле Муавра,
