Задача 41932 ...
Условие
Все решения
a) (3-4i)-(2-i)-(-4+2i)=3-4i-2+i+4-2i=[b]5-5i[/b]
б)
(1-i)^2+i=1-2i+i^2+i=1-2i-1+i=[b]-i[/b]
в)
[m]\frac{(2+4i)(-3-i)}{12-4i}=\frac{-6-12i-2i-4i^2}{12-4i}=\frac{-6-12i-2i+4}{12-4i}=\frac{-2-14i}{12-4i}=\frac{-1-7i}{6-2i}[/m]
[m]=\frac{(-1-7i)(6+2i)}{(6-2i)(6+2i)}=\frac{-6-42i-2i-14i^2}{6^2-(2i)^2)}=\frac{8-44i}{36+4}=\frac{8-44i}{40}=\frac{2-11i}{10}[/m]
г)
(15*(cos12^(o)+isin12^(o)))^5=15^5*(cos(12*5)^(o)+isin(12*5)^(o))=
=(15^5)*(cos60^(o)+isin60^(o))=[m]\frac{(15^{5}\cdot (1+i\sqrt{3})}{2}[/m]
225(сos30^(o)+isin30^(o))=[m]\frac{15^2\cdot(\sqrt(3)+i)}{2}[/m]
Тогда
[m]\frac{(15(cos12^{\circ}+isin12^{\circ}))^{5}}{225(cos30^{\circ}+isin30^{\circ})}=\frac{15^5(cos60^{\circ}+isin60^{\circ})}{15^2(cos30^{\circ}+isin30^{\circ})}=15^3\frac{1+i\sqrt{3}}{\sqrt{3}+i}=[/m]
[m]=\frac{3475(1+i\sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}=\frac{3475(\sqrt{3}+3i-i-\sqrt{3}i^2)}{(\sqrt{3})^2-i^2}=\frac{3475\cdot (2\sqrt{3}+2i)}{4}=\frac{3475\cdot (\sqrt{3}+i)}{2}[/m]
д)
[m](e^{i\cdot \pi })^2=e^{i\cdot 2\cdot\pi}=cos2\pi +isin2\pi =1+i\cdot0=1[/m]
[m](e^{i\cdot \frac{\pi }{2} })^2=e^{i\cdot (\frac{\pi }{2})\cdot 2}=e^{i\pi }=cos\pi +isin\pi =-1=i\cdot0=-1[/m]
1*(-1)=-1
О т в е т. -1