örnek sayfalar - Ezbersiz Matematik

64
ELEMENTLER
01 10  4

01 10  13
y
= 1  x 2  y 2 = 20  x = 4  2
x2
 tan  2  1  =
 y x   y  2 x
=T
1  y x.  y  2  x
1  y x.  y  2  x = 0  x 2  y 2  2 y  0
01 10  5
 z = 1 = z  1  x 2  y 2 = 1   x  1  y 2 = 1
2
01 10  14
1
3
1
3
2 4
 x= ,y=
 z 1=  
i

2
2
2 2
3
3
 k = 4  2  2.2.
2
01 10  6
 tan  =
7.CİLT
2
2
1
k=2 3
2
01 10  15
2 1 4
7
7
= , z = 17. 5  z = 85 c i s Arc tan
1  2. 1 4  6
6
 z = 5  5 3i  10i = 10 c i s 240  10 c i s 90
= 10   cos 240  cos 90    sin 240  sin 90  i ]
7 
 6
= 85 

i  = 6  7i
85 
 85
= 20  cos165 cos 75  i sin165 cos 75  
tan  = tan165   = 165 , II .Yol : Şekil çizerek
01 10  7
 x  2   y  4
 1, 1  d = 5 2
2

2
=
 x  4
2
01 10  16
  y  6  x  y  8 = 0
2
2
01 10  8
k  2   
10
2
2
01 11  1
   =
9.2    18

=
10
10
10
= 18.18 = 324 veya düzgün ongenden  360  36 = 324 
01 10  9
Ahmet YAZICI
 10 z = 10 5 c i s
 1
 2
 k = 3  2  2.3.     k = 19

m 4
 = 1  m.n = 8  m 2  4 2  n 2  2 2 = 32.6
4 2
m 2  n 2 = 34  m  n = 5 2  5 2
01 11  2
y6 y2


  2  1 =  tan  2  1  = x  5 x  1 = T
y6 y2
2
1

x  5 x 1
y6 y2
2
2
 1

  x  2    y  2  = 25
x  5 x 1
 2, 2    7,10   13  5 = 8

 z2 = 8c i s 70

 k
01 11  3


 z  z = c i s120  c i s80 
3
5
2
 cos120  cos 80   sin120  sin 80 

x 2  y 2 = 2  2i  y = 2, x = 0  4  3 =
3
2
01 10  11

 x  u
2
2

= 2 1  cos 80

= 2 cos 40 = 2 cos160 , II .Yol : Şekil çizerek, kos.teo.
01 10  10
 x  yi 
= 82  62  k = 10
2
  y  v =
2
 x  u   y  v
xu  y x  v u 
2


 z1 = 8cis165  z2 = 6cis75  k = 6  8  k = 10
3
2
2
2
01 11  5
2
 x.u  y.v = 0
y xv u

=
= T   2  1 =
y v
x.u  y.v
2
1 .



0
x u
II .Yol : Köşegenleri eş olan paralelkenar dikdörtgendir.
tan  2  1  =
01 11  4
01 10  12
y
3
3
1
3
2

= 3   x  1  y 2 = 1  
i 
i
x 1
2 2
2 2
 5
tan  = 3 3   3   = 
, II .Yol : Şekil çizerek
6 3
12 = 4  2  2.2.4 cos   cos  = 1 2 , z1 z 2 =
2
2
01 11  6
 c i s 30.


3 1 
 i  3  i = 2

3  i  = 
 2
2 
01 11  7





 2 c i s 222 .c i s 42 = 2 c i s180 = 2
1
c i s 60
2


65
KARMAŞIK SAYILAR -ÇÖZÜMLER
01 12  4
01 11  8
 c i s 20. z = 2  i  c i s110. z = x  yi 
x  yi
= c i s 90 = i
2 i
1  2i  1 1  2i  1
=
 i , 1  i 
2
2
01 12  5

x  yi = 1  2i

 P  x =  x  9
01 11  9
  3  i  c i s  = 1  3i  c i s  =
1  3i
= i   = 90
3 i
2
 4   P  0  = 36


  x  2  x  4 x  7 = x  2x  x +14 = 0

 4 2 c i s 225.c i s 75 = 4 2 c i s 300  2 2, 2 6
2
3
2

01 12  7
01 11  11
  = 4  3  i   4m = 0  m = 8  6 i
2

2 5 cos 150  Arg 1  2i   = 2 5  cos cos  sin sin 

3 1 1 2 
tan  = 2  Re  z  = 2 5  

=  3 2
 2 5 2 5
01 12  8
z=
5  12i = x  yi  x 2  y 2 = 5  xy = 6
   3  2i , 3  2i 
01 11  12
1
3

i
2 2
01 11  13
cos 2 2  sin 2 2
1
= =1
2
sin  cos  
 sin   cos 
 1
2
4
cos  sin  
2
2
4
2
01 11  14
01 12  9
Ahmet YAZICI
 x 2  y 2  2ixy  x  yi = 0  x = 1 2  y =  3 2 
2
 x
01 12  6
01 11  10

2
z=
2  2 3i = 4 c i s 60 = 2 c i s 30    3  i , 3  i 
01 12  10
 1  i   1  i   x3 = 0  x3 = 2  2 1  i  = 2  2i
01 12  11
x
 x2  y2 = 4
x y
2
2
 x2  y2  4 x  0
 z 3 = 8c i s180  z1 = 2 c i s 60 = 1  3i  z2 = 1  3i
01 12  12
01 11  15

2
 8 =   k 2   =
7
01 11  16
cos 2 = cos2  


 cos 4 = 1  4 = k 2   = ,  ,
2
sin 2 =  sin 2  
2i  2i  x1  x2 = 4  x1  x2 = 4  2
2
 x1  x2 = 16  10 = 26
2i.2i. x1. x2 = 20  x1. x2 = 5

01 12  13
 x  y  2ixy  4 x  4 yi  4 = 0  x = 2  y  4
2
2
01 12  14
01 12  2
01 12  1
 x  8 = 0  x = 8   
2
2
6  4
= 3 i
2
 z 3 = 64 c i s 0  z = 4 c i s120k  4 c i s 0 , 4 c i s120 , 4 c i s 240

 

 2  2 3i  2  3i = 4
01 12  15
01 12  3
 1  i 1  i  x3 = 8  x3 = 4  64  2.16  4k  8 = 0
k = 6
360k
 2 c i s 0 , 2 c i s 60
6
2 c i s120 , 2 c i s180 , 2 c i s 240 , 2 c i s 300  900 = 5
 z 6 = 64 c i s 0  z = 2 c i s

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