## anonymous 4 years ago HELPP .. 10i/10-i solve in z=x+iy (compleks conjugate)

1. anonymous

you can solve it by rationalizing |dw:1343807584287:dw|

2. anonymous

i dont get it .

3. anonymous

now multiply dominaters and nominators

4. Neo92

10i/10-i =(0+10i)/(10-i) =(0+10i)(10+i)/(10-i)(10+i) do the remaining few calculations and tell me the answer..

5. Neo92

still didnt get it??

6. anonymous

100i + 10i^2/100-i^2 ?

7. anonymous

what should i do to get conjugate form ?

8. anonymous

Denominator part is right but need to solve further.. $100 - i^2 = ??$

9. anonymous

$\large i^2 = -1$

10. anonymous

i need x+iy form.. :( confuse

11. anonymous

Yes you will get that.. Just have patience.. Tell me : $100 - i^2 = ??$

12. anonymous

The value of $$i^2$$ is -1..

13. anonymous

i=10 ?

14. anonymous

or 100+1 = 101 ?

15. anonymous

Last one is right..

16. anonymous

Similarly do it in the Numerator,,,

17. Neo92

@greysica i^2 = -1 therefore 100 - (-1) ==?

18. anonymous

$100i + 10 i^2 = ??$

19. anonymous

100(-1)+10(-1)^2 = -100+10 =-90 ? like this ?

20. Neo92

(100i - 10) use the fact that i^2 = -1

21. anonymous

You can change i^2 there and not i.. remain i as such..

22. anonymous

$\large 100i + 10i^2 = 100i + 10(-1) = ??$

23. anonymous

ohh i see , 100i -10 ?

24. anonymous

Yes...

25. anonymous

Now you have to find this in the form z = x + iy.. So: $\frac{-10 + 100i}{101} \implies \frac{-10}{101} + i \frac{100}{101}$

26. anonymous

Here I have separated both.. does it look like x + iy ??

27. anonymous

yes , i do understand . but my teacher already give me an answer for this , he said it should be 10+i and i must find the way to get 10 + i

28. anonymous

See your teacher said to you want to solve this by using $$10 + i$$ Here we have done the same ..

29. anonymous

The conjugate of $$10 - i$$ is $$10 + i$$ So we have multiplied and divide it by the conjugate that is : $$10 + i$$

30. anonymous

i dont understand . we've got -10/101+i100/101 and my teacher said 10+i as the answer

31. anonymous

This will not be the answer for this.. Ask him again when you will meet him..

32. anonymous

10 + i is the conjugate for this..

33. anonymous

I think he just said to find conjugate so then the answer should be $$10 + i$$

34. anonymous

okay ,how about this.. 11+2i / 4+3i I do some calculation and get 2+i as the answer but my teacher said the asnwer is 2-i

35. anonymous

Let me check..

36. anonymous

2 - i is right..

37. anonymous

okay i will write my calculation ..

38. anonymous

Can you show me..??

39. anonymous

|dw:1343814167411:dw|

40. anonymous

|dw:1343814339078:dw|

41. anonymous

50/25 - i (-25/25) 50/25 + i 2+i

42. anonymous

You are doing mistake of minus.. Let me understand what you did and what the way of your teacher to solve this complex numbers problems..

43. anonymous

|dw:1343814575597:dw|

44. anonymous

|dw:1343814799716:dw|

45. anonymous

Can you once try this ?? Do calculations according to it.. Try once..

46. anonymous

ok wait

47. anonymous

Wait..

48. anonymous

|dw:1343815363469:dw|

49. anonymous

|dw:1343815404009:dw|

50. anonymous

Here x1 = 11, x2 = 4 y1 = 2 and y2 = 3

51. anonymous

yap , the asnwer is 2+i

52. anonymous

2 + i ??

53. anonymous

ohh the formula from my book is wrong , omg ..

54. anonymous

nono i mean 2-i

55. anonymous

Is it clear to you now ??

56. anonymous

See i show you more properly how I do this types of problems..

57. anonymous

$\frac{11 + 2i}{4 + 3i} \times \frac{4 - 3i}{4 - 3i} \implies \frac{44 - 33i + 8i - 6i^2}{(4)^2 - (3i)^2} \implies \frac{44 - 25i -6(-1)}{16-9i^2}$ $\frac{44 - 25i + 6}{16 - 9(-1)} \implies \frac{50 - 25i}{16 + 9} \implies \frac{50 - 25i}{25} \implies \frac{50}{25} - i \frac{25}{25} \implies 2 - i$

58. anonymous

okay , thank you !! if u dont mind , i still have a little confuse , if z1/z2 = x1+iy1/x2+iy2 how about z2/2z1 ? is that means .. wait i draw

59. anonymous

|dw:1343816101532:dw|

60. anonymous

You have to multiply and divide by the conjugate.. Do you know how to find conjugate of a complex number ??

61. anonymous

|dw:1343816319564:dw|

62. anonymous

like that ?

63. anonymous

yes well done.. Need not to multiply 2.. You can simply do this: |dw:1343816404418:dw|

64. anonymous

hmm i see .. i really suck at this actually .. but i try to understand .. wait i try to do some calculation, thank you btw

65. anonymous

66. anonymous

|dw:1343817123248:dw| there is i^2

67. anonymous

|dw:1343817171131:dw| and dissapear in this .. where is i^2 ?

68. anonymous

|dw:1343817264875:dw|

69. anonymous

$\large -y_1y_2(i^2) = -y_1y_2(-1) \implies \color{blue}{+y_1y_2}$

70. anonymous

ohh I understand , so i^2 always meaning -1 in compleks form ?

71. anonymous

Yes..

72. anonymous

Remember: $\large i = \sqrt{-1}$ So need to change $$i$$.. $\large i^2 = -1$ $\large i^3 = -i$ $\large i^4 = 1$

73. anonymous

|dw:1343817500482:dw| is that correct ?