Выполнить действияРешить: а,б,в,г )Во вложении..

Question 1

Выполнить действия
Решить: а,б,в,г )
Во вложении..

Question 2

и пункты д,е ) Забыл дописать.Простите

Question 3

Решите задачу:

$a) \ (3 + 5i) - (6 + 3i) = (3 - 6) + (5 - 3)i = \boxed{-3 + 2i}\\\\ b) \ (5 - 4i)(3 + 2i) = 5*3 + 10i - 12i - 8i^2 =\\\\ = (15 + 8) + (10 - 12)i = \boxed{23 - 2i}\\\\ \left[ \ (a + bi)(c + di) = (ac - bd) + (ad + dc)i \ \right]\\\\$

$c) \ \frac{(3 + 2i)(2 - i)}{(2 + 3i)(1 + i)} = \frac{(6 + 2) + (4 - 3)i}{(2 - 3) + (2 + 3)i} = \frac{8 + i}{-1 + 5i} = \frac{(8 + i)(-1 - 5i)}{(-1 + 5i)(-1 - 5i)} =\\\\ = \frac{(-8 + 5)}{(-1)^2 - (5)^2} + \frac{(-1 - 40)}{(-1)^2 + (5)^2}i = \boxed{-\frac{3}{26} - \frac{41}{26}i}$

$\left[ \ \frac{a + bi}{c + di} = \frac{(ac + bd)}{c^2 + d^2} + \frac{(bc - ad)}{c^2 + d^2}i \ \right]\\\\ d) \ \frac{1 - 3i}{i - 2} + \frac{4i + 1}{3i - 1} = \frac{1 - 3i}{-2 + i} + \frac{1 + 4i}{-1 + 3i} = \frac{-2 - 3}{1 + 2^2} + \frac{6 - 1}{1 + 2^2}i + \frac{-1 + 12}{1 + 3^2} + \frac{-4 - 3}{1 + 3^2}i =\\\\= -1 + i + \frac{11}{10} - \frac{7}{10}i = ( \frac{11}{10} - 1) + (1 - \frac{7}{10})i = \boxed{\frac{1}{10} + \frac{3}{10}i}$

$e) \ (\frac{\sqrt{2}}{2} + \frac{1}{2}i)^2 = (\frac{\sqrt{2}}{2} + \frac{1}{2}i) (\frac{\sqrt{2}}{2} + \frac{1}{2}i) = \\\\ = ((\frac{\sqrt{2}}{2})^2 - (\frac{1}{2})^2) + (\frac{\sqrt{2}}{2}\frac{1}{2} + \frac{\sqrt{2}}{2}\frac{1}{2})i = \boxed{\frac{1}{4} + \frac{\sqrt{2}}{2}i}$

$f) \ \frac{2 + 3i}{(4 + i)(2 + 2i)} = \frac{2 + 3i}{(8 - 2) + (2 + 8)i} = \frac{2 + 3i}{6 + 10i} = \frac{12 + 30}{36+ 100} + \frac{18 - 20}{36+ 100}i = \boxed{\frac{21}{68} - \frac{1}{68}i}$