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мохинсан_zn · Answer 1 · 2018-03-16T09:47:16+0000

1.
0} \atop {\sin\phi<0}} \right.===>\phi\in(-\frac{3\pi}{2}+2\pi n;2\pi+2\pi n) ==>\\ ==>\phi=-\frac\pi4+2\pi n=\frac{7\pi}{4}+2\pi n\\ z=\sqrt2\left(\cos(\frac{7\pi}{4}+2\pi n)+i\sin(\frac{7\pi}{4}+2\pi n)\right);\\ \\ z=\sqrt2\cdot e^{i(\frac{7\pi}{4}+2\pi n)}" alt="z=1-i;\\ z=x+iy;\\ z=\rho\cdot(\cos\phi+i\sin\phi); \rho=\sqrt{x^2+y^2};\\ \phi=arctg\frac y x;\\ x=1;\ \ y=-1;\\ \rho=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt2;\\ \phi=arctg(\frac{-1}{1})+\pi n=arctg(-1)+\pi n=-\frac{\pi}{4}+\pi n;\\ \left \{ {{\cos\phi>0} \atop {\sin\phi<0}} \right.===>\phi\in(-\frac{3\pi}{2}+2\pi n;2\pi+2\pi n) ==>\\ ==>\phi=-\frac\pi4+2\pi n=\frac{7\pi}{4}+2\pi n\\ z=\sqrt2\left(\cos(\frac{7\pi}{4}+2\pi n)+i\sin(\frac{7\pi}{4}+2\pi n)\right);\\ \\ z=\sqrt2\cdot e^{i(\frac{7\pi}{4}+2\pi n)}" align="absmiddle" class="latex-formula">
2.
$z=\frac{i}{(1+i)^2};\\ z=\frac{1}{(1+i)^2}=\frac{i\cdot(1-i)^2}{(1+i)^2(1-i)^2}=\frac{i\cdot(1-2i+(-1))}{(1-(-1))^2}=\frac{i\cdot(-2i)}{2^2}=\frac{2-}{4}=\frac{1}{2}=\\ =\frac12+i\cdot0\\ \frac{i}{(1+i)^2}=\frac{i}{1+2i+i^2}=\frac{i}{1+2i-1}=\frac{i}{2i}=\frac12=\frac12+0\cdot i\\ x=\frac12;\ \ y=0;\\ \rho=\sqrt{x^2+y^2}=\sqrt{(\frac12)^2+0^2}=\sqrt{\frac14}=\frac12;\\ \phi=arctg{\frac{0}{\frac12}}=0;\\ \phi=2\pi n,\ n\in Z;\\ z=\frac12(\cos2\pi n+i\sin2\pi n)\\ z=\frac12\cdot e^{i2\pi n}$
3.
$(\sqrt3-i)^9=\\ z=\sqrt3-i;\ \ z^9-?\\ x=\sqrt3; y=-1\\ \rho=\sqrt{(\sqrt3)^2+(-1)^2}=\sqrt{3+1}=\sqrt4=2;\\ \phi=arctg\frac y x=arctg\frac{-1}{\sqrt3}=-\frac\pi6+2\pi n=\frac{11\pi}{6}+2\pi n;\\ z=2\cdot e^{i(\frac{11\pi}{6}+2\pi n)};\\ z^9=2^9\cdot e^{9i(\frac{11\pi}{6}+2\pi n)}=512\cdot e^{\frac{99\pi}{6}+18\pi n}=\\ =512\cdot(\cos(\frac{99\pi}{6}+18\pi n)+i\sin(\frac{99\pi}{6}+18\pi n))=\\ |\frac{99\pi}{6}+2\pi n=\frac{96\pi}{6}+\frac{3\pi}{6}+2\pi n=16\pi +\frac\pi2+2\pi n=\frac\pi2+2\pi k\\$
$z^9=512\cdot\left(\cos(\frac\pi2+2\pi n)+i\sin(\frac\pi2+2\pi n)\right)=512(0+1\cdo i)=512i;\\ (\sqrt3-i)^9=512i;\\$
4.
$\frac{(-\sqrt3+i)(\cos\frac\pi{12}-i\sin\frac\pi{12})}{1-i}=\frac{(-\sqrt3+i)(\cos\frac\pi{12}-i\sin\frac\pi{12})(1+i)}{(1-i)(1+i)}=\\ =\frac{(-\sqrt3-\sqrt3i+i-1)(\cos\frac{\pi}{12}-i\sin\frac{\pi}{12})}{1+1}=\\ |\cos\frac{\pi}{12}-i\sin\frac{\pi}{12}=\cos(-\frac{\pi}{12})+i\sin(-\frac{\pi}{12})=e^{-\frac{\pi}{12}}|\\ =\frac{(-(\sqrt3+1)-i(\sqrt3-1))\cdot e^{-\frac{\pi}{12}}}{2}=\\$
$=-\frac{(\sqrt3+1)\cos\frac{\pi}{12}+(\sqrt3-1)\sin\frac{\pi}{12}}{2}+i\frac{(\sqrt3-1)\sin\frac\pi{12}-(\sqrt3-1)\cos\frac\pi{12}}{2}$