Задание 2.
В задании требуется
определить все характеристики двух
комплексных чисел
и
результата выполнения над ними
арифметической операции.
Таблица 1. Характеристики комплексных чисел.
|
№ вар. |
Число z1 |
Число z2 |
Число z (результат) | ||||||||||
|
Вещ |
Мн. |
Мод. |
Арг. |
Вещ. |
Мн. |
Мод. |
Арг. |
Опе ра- ция |
Вещ. |
Мн. |
Мод. |
Арг. | |
|
1 |
2 |
– |
3 |
– |
– |
– |
5 |
–300 |
+ |
– |
– |
– |
– |
|
2 |
– |
– |
5 |
00 |
–2 |
–6 |
– |
– |
– |
– |
– |
– |
– |
|
3 |
4 |
|
– |
450 |
– |
– |
6 |
1500 |
/ |
– |
– |
– |
– |
|
4 |
– |
– |
2 |
1350 |
–1 |
–1 |
– |
– |
* |
– |
– |
– |
– |
|
5 |
|
2 |
4 |
– |
– |
– |
4 |
700 |
+ |
– |
– |
– |
– |
|
6 |
– |
– |
3,5 |
600 |
–1 |
–1 |
– |
– |
– |
– |
– |
– |
– |
|
7 |
5 |
|
– |
2700 |
– |
– |
8 |
1200 |
/ |
– |
– |
– |
– |
|
8 |
– |
– |
3 |
600 |
0 |
5 |
– |
– |
* |
– |
– |
– |
– |
|
9 |
1 |
|
1 |
– |
– |
– |
3 |
–600 |
+ |
– |
– |
– |
– |
|
10 |
– |
– |
2 |
450 |
3 |
0 |
– |
– |
+ |
– |
– |
– |
– |
|
11 |
|
–7 |
7 |
– |
– |
– |
2 |
550 |
– |
– |
– |
– |
– |
|
12 |
– |
– |
5 |
450 |
3 |
1 |
– |
– |
/ |
– |
– |
– |
– |
|
13 |
7 |
|
– |
1800 |
– |
– |
4 |
300 |
* |
– |
– |
– |
– |
|
14 |
– |
– |
1 |
900 |
|
|
– |
– |
+ |
– |
– |
– |
– |
|
15 |
–5 |
–5 |
– |
– |
– |
– |
2 |
00 |
– |
– |
– |
– |
– |
|
16 |
– |
– |
3 |
1800 |
2 |
–2 |
– |
– |
/ |
– |
– |
– |
– |
|
17 |
–2 |
2 |
– |
– |
– |
– |
5 |
900 |
* |
– |
– |
– |
– |
|
18 |
– |
– |
4 |
900 |
3 |
–3 |
– |
– |
+ |
– |
– |
– |
– |
|
19 |
|
-7 |
10 |
– |
– |
– |
7 |
-900 |
– |
– |
– |
– |
– |
|
20 |
– |
– |
2 |
300 |
9 |
2 |
– |
– |
/ |
– |
– |
– |
– |
|
21 |
2 |
|
5 |
– |
– |
– |
6 |
1800 |
* |
– |
– |
– |
– |
|
22 |
– |
– |
3 |
-300 |
0,5 |
0,2 |
– |
– |
+ |
– |
– |
– |
– |
|
23 |
3 |
2 |
– |
– |
6 |
– |
– |
1350 |
– |
– |
– |
– |
– |
|
24 |
– |
– |
6 |
–600 |
2 |
–3 |
– |
– |
/ |
– |
– |
– |
– |
|
25 |
3 |
–3 |
– |
– |
– |
– |
3 |
450 |
* |
– |
– |
– |
– |
|
26 |
2 |
5 |
– |
– |
– |
– |
6 |
1800 |
* |
– |
– |
– |
– |
|
27 |
– |
– |
3 |
-900 |
0,5 |
0,25 |
– |
– |
+ |
– |
– |
– |
– |
|
28 |
3 |
2 |
– |
– |
6 |
– |
– |
1800 |
+ |
– |
– |
– |
– |
|
29 |
– |
– |
6 |
–600 |
2 |
–3 |
– |
– |
/ |
– |
– |
– |
– |
|
30 |
4 |
–4 |
– |
– |
– |
– |
3 |
450 |
* |
– |
– |
– |
– |
Задание 3.
В
соответствии с заданием, приведенным
в табл. 2, необходимо на одном двумерном
графике построить графики двух функций
,
трассировкой графиков найти координаты
точек пересечения, построить трехмерный
график и выбрать наиболее информативный
ракурс его представления.
Таблица 2. Формулы для функций одной и двух переменных и интервалы изменения аргументов.
|
№ вар |
|
|
Ин-тер- вал x |
|
Интер- валы x,y |
|
1 |
|
|
[0,4] |
|
[0,10], [0,5] |
|
2 |
|
|
[-1,6] |
|
[0,10], [0,15] |
|
3 |
|
|
[0,5] |
|
[-10,10], [-10,10] |
|
4 |
|
|
[0,2] |
|
[0,10], [0,10] |
|
5 |
|
|
[0,8] |
|
[0,4], [0,3] |
|
6 |
|
|
[1,5] |
|
[0,3], [0,3] |
|
7 |
|
|
[1,4] |
|
[0,4], [0,4] |
|
8 |
|
|
[0,4] |
|
[0,15], [0,5] |
|
9 |
|
|
[0,4] |
|
[0,15], [0,20] |
|
10 |
|
|
[-1,1] |
|
[0,15], [0,10] |
|
11 |
|
|
[-1,4] |
|
[0,15], [0,10] |
|
12 |
|
|
[1,4] |
|
[0,25], [0,25] |
|
13 |
|
|
[0,4] |
|
[0,10], [0,5] |
|
14 |
|
|
[-1,6] |
|
[0,10], [0,15] |
|
15 |
|
|
[0,5] |
|
[-10,10], [-10,10] |
|
16 |
|
|
[0,2] |
|
[0,10], [0,10] |
|
17 |
|
|
[0,8] |
|
[0,4], [0,3] |
|
18 |
|
|
[1,5] |
|
[0,3], [0,3] |
|
19 |
|
|
[1,4] |
|
[0,4], [0,4] |
|
20 |
|
|
[0,4] |
|
[0,15], [0,5] |
|
21 |
|
|
[0,4] |
|
[0,10], [0,5] |
|
22 |
|
|
[-1,4] |
|
[0,15], [0,10] |
|
23 |
|
|
[1,4] |
|
[0,25], [0,25] |
|
24 |
|
|
[0,4] |
|
[0,10], [0,5] |
|
25 |
|
|
[-1,6] |
|
[0,10], [0,15] |
|
26 |
|
|
[-1,1] |
|
[0,15], [0,10] |
|
27 |
|
|
[-1,4] |
|
[0,15], [0,10] |
|
28 |
|
|
[1,4] |
|
[0,25], [0,25] |
|
29 |
|
|
[0,4] |
|
[0,10], [0,5] |
|
30 |
|
|
[-1,6] |
|
[0,10], [0,15] |
