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

недорешали, не найдены корни из комплексных чисел

Answer 3

Было задание представить в тригонометрической форме

Answer 4

да, представить в триг .форме квадратные корни из комплексных чисел

Answer 5

$1); ; omega =sqrt{sqrt3+i}\\z=sqrt3+i; ; to ; ; x=sqrt3>0; ,; y=1>0; ; to \\r=|z|=sqrt{x^2+y^2}=sqrt{3+1}=2; ,\\cosphi =frac{x}{sqrt{x^2+y^2}}=frac{sqrt3}{2}>0; ,; ; sinphi =frac{y}{sqrt{x^2+y^2}}=frac{1}{2}>0; ,; -pi leq phi leq pi ; Rightarrow \\tgphi =frac{sinphi }{cosphi }=frac{1}{sqrt3}=frac{sqrt3}{3}; ,; ; phi =arctgfrac{y}{x}=arctgfrac{sqrt3}{3}=frac{pi}{6}; .\\z=rcdot (cosphi+i, sinphi )=2cdot (cosfrac{pi }{6}+i, sinfrac{pi}{6})$

$star ; sqrt[n]{z}=sqrt[n]{r}cdot (cos, frac{phi +2pi k}{n}+icdot sin, frac{phi +2pi k}{n}); ,; -pi leq phi leq pi ; ,; k=0,1,...,n-1.\\omega =sqrt{z}=sqrt{sqrt3+i}\\omega _0=sqrt2cdot (cosfrac{pi /6+2pi cdot 0}{2}+i, sinfrac{pi /6+2pi cdot 0}{2})=sqrt2cdot (cosfrac{pi }{12}+i, sinfrac{pi }{12})\\omega_1=sqrt2cdot (cosfrac{pi /6+2pi cdot 1}{2}+i, sinfrac{pi /6+2pi cdot 1}{2})=sqrt2cdot (cosfrac{13pi }{12}+i, sinfrac{13pi }{12})$

$2); ; omega =sqrt{1-i}\\z=1-i; ; Rightarrow ; ; x=1>0; ,; y=-1<0; ; Rightarrow ; ; r=|z|=sqrt{1+1}=sqrt2\\tgphi =arctg(-1)=-arctg1=-frac{pi}{4}\\z=sqrt2cdot (cos(-frac{pi}{4})+i, sin(-frac{pi}{4}))\\omega=sqrt{z}=sqrt{1-i}\\omega_0=sqrt{sqrt2}cdot (cosfrac{-pi /4}{2}+i, sinfrac{-pi /4}{2})=sqrt[4]2cdot (cos(-frac{pi }{8})+i, sin(-frac{pi}{8}))\\omega _1=sqrt[4]2cdot (cosfrac{-pi /4+2pi }{2}+i, sinfrac{-pi /4+2pi }{2})=sqrt[4]2cdot (cosfrac{7pi}{8}+i, sinfrac{7pi}{8})$

$3); ; z=1+2i; ,; ; z^3=(1+2i)^3\\x=1>0; ,; ; y=2>0; ,; ; tgphi =frac{y}{x}=2; ,; phi =arctg2in [-pi ,pi ]\\r=|z|=sqrt{x^2+y^2}=sqrt{1+4}=sqrt5\\z=sqrt5cdot (cos(arctg2)+i, sin(arctg2))\\star ; ; z^{n}=r^{n}cdot (cos(nphi )+i, sin(nphi )); ; star \\z^3=sqrt{5^3}cdot (cos(3, arctg2)+i, sin(3arctg2))$

$4); ; omega =sqrt{1+i}; ; to ; z=1+i\\x=1>0; ,; y=1>0; ; to ; ; phi =arctgfrac{y}{x}=arctg1=frac{pi }{4}\\r=|z|=sqrt{x^2+y^2}=sqrt{1+1}=sqrt2\\z=sqrt2cdot (cosfrac{pi }{4}+i, sinfrac{pi }{4})\\omega =sqrt{z}=sqrt{1+i}\\omega_0=sqrt{sqrt2}cdot (cosfrac{pi /4}{2}+i, sinfrac{pi /4}{2})=sqrt[4]2cdot (cosfrac{pi}{8}+i, sinfrac{pi}{8})\\omega _1=sqrt[4]2cdot (cosfrac{pi /4+2pi }{2}+i, sinfrac{pi /4+2pi }{2})=sqrt[4]2cdot (cosfrac{9pi}{8}+i, sinfrac{9pi}{8})$

$5); ; omega =sqrt{1-2i}; ; to ; ; z=1-2i\\x=1>0; ,; ; y=-2<0; ; to ; ; r=|z|=sqrt{1+4}=sqrt5\\phi =arctgfrac{-2}{1}=-arctg2in [-pi ;pi ]\\z=sqrt5cdot (cos(-arctg2)+i, sin(-arctg2))\\omega =sqrt{z}=sqrt{1-2i}\\omega _0=sqrt{sqrt5}cdot (cosfrac{-arctg2}{2}+i, sinfrac{-arctg2}{2})=sqrt[4]5cdot (cos(-frac{arctg2}{2})+i, sin(-frac{arctg2}{2}))\\omega _1=sqrt[4]5cdot (cosfrac{2pi -arctg2}{2}+i, sinfrac{2pi -arctg2}{2})$

$6); ; omega =sqrt{1-sqrt3i}; ; to ; ; z=1-sqrt3i\\x=1>0; ,; y=-sqrt3<0; to ; ; r=|z|=sqrt{1+3}=2\\phi =arctgfrac{-sqrt3}{1}=-arctgsqrt3=-frac{pi}{3}in [-pi ;pi ]\\z=2cdot (cos(-frac{pi}{3})+i, sin(-frac{pi}{3}))\\omega =sqrt{z}=sqrt{1-sqrt3i}\\omega _0=sqrt2cdot (cosfrac{-pi /3}{2}+i, sinfrac{-pi /3}{2})=sqrt2cdot (cos(-frac{pi}{6})+i, sin(-frac{pi}{6}))\\omega _1=sqrt2cdot (cosfrac{-pi /3+2pi }{2}+i, sinfrac{-pi /3+2pi }{2})=sqrt2cdot (cosfrac{5pi}{6}+i, sinfrac{5pi }{6})$

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