Question

Решите пожалуйста!!!

Answer 1

1.
$z=1-i;\ z=x+iy;\ z=rhocdot(cosphi+isinphi); rho=sqrt{x^2+y^2};\ phi=arctgfrac y x;\ x=1; y=-1;\ rho=sqrt{1^2+(-1)^2}=sqrt{1+1}=sqrt2;\ phi=arctg(frac{-1}{1})+pi n=arctg(-1)+pi n=-frac{pi}{4}+pi n;\ left { {{cosphi>0} atop {sinphi<0}} right.===>phiin(-frac{3pi}{2}+2pi n;2pi+2pi n) ==>\ ==>phi=-fracpi4+2pi n=frac{7pi}{4}+2pi n\ z=sqrt2left(cos(frac{7pi}{4}+2pi n)+isin(frac{7pi}{4}+2pi n)right);\ \ z=sqrt2cdot e^{i(frac{7pi}{4}+2pi n)}$
2.
$z=frac{i}{(1+i)^2};\ z=frac{1}{(1+i)^2}=frac{icdot(1-i)^2}{(1+i)^2(1-i)^2}=frac{icdot(1-2i+(-1))}{(1-(-1))^2}=frac{icdot(-2i)}{2^2}=frac{2-}{4}=frac{1}{2}=\ =frac12+icdot0\ frac{i}{(1+i)^2}=frac{i}{1+2i+i^2}=frac{i}{1+2i-1}=frac{i}{2i}=frac12=frac12+0cdot i\ x=frac12; y=0;\ rho=sqrt{x^2+y^2}=sqrt{(frac12)^2+0^2}=sqrt{frac14}=frac12;\ phi=arctg{frac{0}{frac12}}=0;\ phi=2pi n, nin Z;\ z=frac12(cos2pi n+isin2pi n)\ z=frac12cdot e^{i2pi n}$
3.
$(sqrt3-i)^9=\ z=sqrt3-i; z^9-?\ x=sqrt3; y=-1\ rho=sqrt{(sqrt3)^2+(-1)^2}=sqrt{3+1}=sqrt4=2;\ phi=arctgfrac y x=arctgfrac{-1}{sqrt3}=-fracpi6+2pi n=frac{11pi}{6}+2pi n;\ z=2cdot e^{i(frac{11pi}{6}+2pi n)};\ z^9=2^9cdot e^{9i(frac{11pi}{6}+2pi n)}=512cdot e^{frac{99pi}{6}+18pi n}=\ =512cdot(cos(frac{99pi}{6}+18pi n)+isin(frac{99pi}{6}+18pi n))=\ |frac{99pi}{6}+2pi n=frac{96pi}{6}+frac{3pi}{6}+2pi n=16pi +fracpi2+2pi n=fracpi2+2pi k$
$z^9=512cdotleft(cos(fracpi2+2pi n)+isin(fracpi2+2pi n)right)=512(0+1cdo i)=512i;\ (sqrt3-i)^9=512i;$
4.
$frac{(-sqrt3+i)(cosfracpi{12}-isinfracpi{12})}{1-i}=frac{(-sqrt3+i)(cosfracpi{12}-isinfracpi{12})(1+i)}{(1-i)(1+i)}=\ =frac{(-sqrt3-sqrt3i+i-1)(cosfrac{pi}{12}-isinfrac{pi}{12})}{1+1}=\ |cosfrac{pi}{12}-isinfrac{pi}{12}=cos(-frac{pi}{12})+isin(-frac{pi}{12})=e^{-frac{pi}{12}}|\ =frac{(-(sqrt3+1)-i(sqrt3-1))cdot e^{-frac{pi}{12}}}{2}=$
$=-frac{(sqrt3+1)cosfrac{pi}{12}+(sqrt3-1)sinfrac{pi}{12}}{2}+ifrac{(sqrt3-1)sinfracpi{12}-(sqrt3-1)cosfracpi{12}}{2}$

Answer 2

мы такого ещё даже не проходили