Question

$\sqrt{ \sqrt{3} + i } \\ \sqrt{1 - i} \\ (1 + 2i {)}^{3} \\ \sqrt{1 + i} \\ \sqrt{1 - 2i} \\ \sqrt{1 - \sqrt{3i} } \\ \sqrt{i}$
комплексные числа

Answer 1

Решение во вложении:

Answer 2

$1)\; \; \omega =\sqrt{\sqrt3+i}\\\\z=\sqrt3+i\; \; \to \; \; x=\sqrt3>0\; ,\; y=1>0\; \; \to \\\\r=|z|=\sqrt{x^2+y^2}=\sqrt{3+1}=2\; ,\\\\cos\phi =\frac{x}{\sqrt{x^2+y^2}}=\frac{\sqrt3}{2}>0\; ,\; \; sin\phi =\frac{y}{\sqrt{x^2+y^2}}=\frac{1}{2}>0\; ,\; -\pi \leq \phi \leq \pi \; \Rightarrow \\\\tg\phi =\frac{sin\phi }{cos\phi }=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}\; ,\; \; \phi =arctg\frac{y}{x}=arctg\frac{\sqrt3}{3}=\frac{\pi}{6}\; .\\\\z=r\cdot (cos\phi+i\, sin\phi )=2\cdot (cos\frac{\pi }{6}+i\, sin\frac{\pi}{6})$

$\star \; \sqrt[n]{z}=\sqrt[n]{r}\cdot (cos\, \frac{\phi +2\pi k}{n}+i\cdot sin\, \frac{\phi +2\pi k}{n})\; ,\; -\pi \leq \phi \leq \pi \; ,\; k=0,1,...,n-1.\\\\\omega =\sqrt{z}=\sqrt{\sqrt3+i}\\\\\omega _0=\sqrt2\cdot (cos\frac{\pi /6+2\pi \cdot 0}{2}+i\, sin\frac{\pi /6+2\pi \cdot 0}{2})=\sqrt2\cdot (cos\frac{\pi }{12}+i\, sin\frac{\pi }{12})\\\\\omega_1=\sqrt2\cdot (cos\frac{\pi /6+2\pi \cdot 1}{2}+i\, sin\frac{\pi /6+2\pi \cdot 1}{2})=\sqrt2\cdot (cos\frac{13\pi }{12}+i\, sin\frac{13\pi }{12})$

$2)\; \; \omega =\sqrt{1-i}\\\\z=1-i\; \; \Rightarrow \; \; x=1>0\; ,\; y=-1<0\; \; \Rightarrow \; \; r=|z|=\sqrt{1+1}=\sqrt2\\\\tg\phi =arctg(-1)=-arctg1=-\frac{\pi}{4}\\\\z=\sqrt2\cdot (cos(-\frac{\pi}{4})+i\, sin(-\frac{\pi}{4}))\\\\\omega=\sqrt{z}=\sqrt{1-i}\\\\\omega_0=\sqrt{\sqrt2}\cdot (cos\frac{-\pi /4}{2}+i\, sin\frac{-\pi /4}{2})=\sqrt[4]2\cdot (cos(-\frac{\pi }{8})+i\, sin(-\frac{\pi}{8}))\\\\\omega _1=\sqrt[4]2\cdot (cos\frac{-\pi /4+2\pi }{2}+i\, sin\frac{-\pi /4+2\pi }{2})=\sqrt[4]2\cdot (cos\frac{7\pi}{8}+i\, sin\frac{7\pi}{8})$

$3)\; \; z=1+2i\; ,\; \; z^3=(1+2i)^3\\\\x=1>0\; ,\; \; y=2>0\; ,\; \; tg\phi =\frac{y}{x}=2\; ,\; \phi =arctg2\in [-\pi ,\pi ]\\\\r=|z|=\sqrt{x^2+y^2}=\sqrt{1+4}=\sqrt5\\\\z=\sqrt5\cdot (cos(arctg2)+i\, sin(arctg2))\\\\\star \; \; z^{n}=r^{n}\cdot (cos(n\phi )+i\, sin(n\phi ))\; \; \star \\\\z^3=\sqrt{5^3}\cdot (cos(3\, arctg2)+i\, sin(3arctg2))$

$4)\; \; \omega =\sqrt{1+i}\; \; \to \; \ z=1+i\\\\x=1>0\; ,\; y=1>0\; \; \to \; \; \phi =arctg\frac{y}{x}=arctg1=\frac{\pi }{4}\\\\r=|z|=\sqrt{x^2+y^2}=\sqrt{1+1}=\sqrt2\\\\z=\sqrt2\cdot (cos\frac{\pi }{4}+i\, sin\frac{\pi }{4})\\\\\omega =\sqrt{z}=\sqrt{1+i}\\\\\omega_0=\sqrt{\sqrt2}\cdot (cos\frac{\pi /4}{2}+i\, sin\frac{\pi /4}{2})=\sqrt[4]2\cdot (cos\frac{\pi}{8}+i\, sin\frac{\pi}{8})\\\\\omega _1=\sqrt[4]2\cdot (cos\frac{\pi /4+2\pi }{2}+i\, sin\frac{\pi /4+2\pi }{2})=\sqrt[4]2\cdot (cos\frac{9\pi}{8}+i\, sin\frac{9\pi}{8})$

$5)\; \; \omega =\sqrt{1-2i}\; \; \to \; \; z=1-2i\\\\x=1>0\; ,\; \; y=-2<0\; \; \to \; \; r=|z|=\sqrt{1+4}=\sqrt5\\\\\phi =arctg\frac{-2}{1}=-arctg2\in [-\pi ;\pi ]\\\\z=\sqrt5\cdot (cos(-arctg2)+i\, sin(-arctg2))\\\\\omega =\sqrt{z}=\sqrt{1-2i}\\\\\omega _0=\sqrt{\sqrt5}\cdot (cos\frac{-arctg2}{2}+i\, sin\frac{-arctg2}{2})=\sqrt[4]5\cdot (cos(-\frac{arctg2}{2})+i\, sin(-\frac{arctg2}{2}))\\\\\omega _1=\sqrt[4]5\cdot (cos\frac{2\pi -arctg2}{2}+i\, sin\frac{2\pi -arctg2}{2})$

$6)\; \; \omega =\sqrt{1-\sqrt3i}\; \; \to \; \; z=1-\sqrt3i\\\\x=1>0\; ,\; y=-\sqrt3<0\; \to \; \; r=|z|=\sqrt{1+3}=2\\\\\phi =arctg\frac{-\sqrt3}{1}=-arctg\sqrt3=-\frac{\pi}{3}\in [-\pi ;\pi ]\\\\z=2\cdot (cos(-\frac{\pi}{3})+i\, sin(-\frac{\pi}{3}))\\\\\omega =\sqrt{z}=\sqrt{1-\sqrt3i}\\\\\omega _0=\sqrt2\cdot (cos\frac{-\pi /3}{2}+i\, sin\frac{-\pi /3}{2})=\sqrt2\cdot (cos(-\frac{\pi}{6})+i\, sin(-\frac{\pi}{6}))\\\\\omega _1=\sqrt2\cdot (cos\frac{-\pi /3+2\pi }{2}+i\, sin\frac{-\pi /3+2\pi }{2})=\sqrt2\cdot (cos\frac{5\pi}{6}+i\, sin\frac{5\pi }{6})$

$7)\; \; \omega=\sqrt{i}\; \; \to \; \; z=i=cos\frac{\pi }{2}+i\, sin\frac{\pi }{2}\\\\\omega _0=cos\frac{\pi}{4}+i\, sin\frac{\pi }{4}\\\\\omega _1=cos\frac{5\pi }{4}+i\, sin\frac{5\pi }{4}$