Помогите решить задание по математике. 10-11 класс 8,9 задание.

Question 1

Question 2

$8. \ z_{1} = 3 - 5i, \ z_{2} = 0,3 + 0,2i, \ z_{3} = \dfrac{1}{4} - \dfrac{1}{2} i, \ z_{4} = -0,4 + 3i$

$z_{1} - z_{2} = (3 - 0,3) + i(-5 - 0,2) = 2,7 - 5,2i$

$z_{1} + z_{3} = \left(3 + \dfrac{1}{4} \right) + i\left(-5 - \dfrac{1}{2} \right) = 3\dfrac{1}{4} - 5\dfrac{1}{2} i$

$z_{2} \cdot z_{3} = \left(0,3 \cdot \dfrac{1}{4} - 0,2 \cdot \left(-\dfrac{1}{2} \right) \right) + i \left(0,3 \cdot \left(-\dfrac{1}{2} \right) + \dfrac{1}{4} \cdot 0,2 \right) =$

$= \dfrac{7}{40} - \dfrac{1}{10}i$

$\dfrac{z_{3}}{z_{4}} = \dfrac{\dfrac{1}{4} \cdot (-0,4) + \left(-\dfrac{1}{2} \right)\cdot 3 }{(-0,4)^{2} + 3^{2}} + i \cdot \dfrac{-0,4 \cdot \left(-\dfrac{1}{2} \right) - \dfrac{1}{4} \cdot 3}{(-0,4)^{2} + 3^{2}} = -\dfrac{40}{229} - \dfrac{55}{916}i$

Действия над комплексными числами в алгебраической форме:

$z_{1} + z_{2} = (x_{1} + iy_{1}) + (x_{2} + iy_{2}) = (x_{1} + x_{2}) + i(y_{1} + y_{2});$

$z_{1} - z_{2} = (x_{1} + iy_{1}) - (x_{2} + iy_{2}) = (x_{1} - x_{2}) + i(y_{1} - y_{2});$

$z_{1} \cdot z_{2} = (x_{1} + iy_{1}) \cdot (x_{2} + iy_{2}) = (x_{1}x_{2} - y_{1}y_{2}) + i(x_{1}y_{2} + x_{2}y_{1});$

$\dfrac{z_{1}}{z_{2}} = \dfrac{x_{1} + iy_{1}}{x_{2} + iy_{2}} = \dfrac{z_{1} \cdot \overline{z}_{2}}{z_{2} \cdot \overline{z}_{2}} =\dfrac{(x_{1} + iy_{1})(x_{2} - iy_{2})}{(x_{2} + iy_{2})(x_{2} - iy_{2})} = \\=\dfrac{x_{1}x_{2} + y_{1}y_{2}}{x_{2}^{2} + y_{2}^{2}} + i \cdot \dfrac{x_{2}y_{1} - x_{1}y_{2}}{x_{2}^{2} + y_{2}^{2}}, \ z_{2} \neq 0.$

Здесь $z_{1} = x_{1} + iy_{1}, \ z_{2} = x_{2} + iy_{2}$

$9. \ z_{1} = 2\sqrt{3} + 2i, \ z_{2} = 6 -6i, \ z_{3} = -\dfrac{\sqrt{3}}{2} - 1,5i, \ z_{4} = 3$

Здесь:

$z_{1} = 4\left(\dfrac{\sqrt{3} }{2} + \dfrac{1}{2} i \right) = 4\left(\cos \dfrac{\pi}{6} + i\sin \dfrac{\pi}{6} \right) \\\ z_{2} = -6\sqrt{2} \left( -\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} i \right) = -6\sqrt{2}\left(\cos \dfrac{3\pi}{4} + i\sin \dfrac{3\pi}{4} \right) \ \\z_{3} = -\sqrt{3} \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) = -\sqrt{3}\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3} \right) \ \\z_{4} = 3(\cos 0 + i\sin 0)$

$z_{4} \cdot z_{3} = 3 \cdot (-\sqrt{3}) \cdot \left[\cos \left(0 + \dfrac{\pi}{3} \right)+ i \sin \left(0 + \dfrac{\pi}{3} \right) \right] = -3\sqrt{3}\left(\dfrac{1}{2} + \dfrac{ \sqrt{3}}{2}i \right) =$

$= -\dfrac{3\sqrt{3}}{2} - \dfrac{9}{2} i$

$\dfrac{z_{4}}{z_{2}} = \dfrac{3}{-6\sqrt{2}} \left[\cos \left(0 - \dfrac{3\pi}{4} \right) + i\sin \left(0 - \dfrac{3\pi}{4} \right) \right] = -\dfrac{\sqrt{2}}{4} \left(-\dfrac{\sqrt{2}}{2} -\dfrac{\sqrt{2}}{2}i \right) =$

$= \dfrac{1}{4} + \dfrac{1}{4} i$

$z_{1}^{2} = 4^{2} \left(\cos \left(2 \cdot \dfrac{\pi}{6} \right) + i\sin \left(2 \cdot \dfrac{\pi}{6} \right)\right) = 16\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3} \right) =$

$= 16 \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i \right) = 8 + 8\sqrt{3}i$

Действия над комплексными числами в тригонометрической форме:

$z_{1} \cdot z_{2} = r_{1}r_{2}\left[\cos(\varphi_{1} + \varphi_{2}) + i\sin(\varphi_{1} + \varphi_{2}) \right];$

$\dfrac{z_{1}}{z_{2}} = \dfrac{r_{1}}{r_{2}} \left[\cos(\varphi_{1} - \varphi_{2}) + i\sin(\varphi_{1} - \varphi_{2}) \right], \ z_{2} \neq 0;$

$z^{n} = \left(r(\cos \varphi + i\sin \varphi \right))^{n} = r^{n}(\cos n \varphi + i\sin n \varphi);$

$\sqrt[n]{z} = \sqrt[n]{r}\left(\cos \dfrac{\varphi + 2\pi k}{n} + i\sin \dfrac{\varphi + 2\pi k}{n} \right), \ k = \overline{0, \ n-1}.$

Здесь $z_{1} = r_{1} (\cos \varphi_{1} + i\sin \varphi_{1}), \ z_{2} = r_{2} (\cos \varphi_{2} + i\sin \varphi_{2}), \ z = r(\cos \varphi + i\sin \varphi).$