## anonymous 5 years ago complex numbers: find a, b such that (-bi)^2 = a^2 ?

1. anonymous

a^2+b^2=0; a=m+ni b=p+qi (m+ni)^2+(p+qi)^2=0 m^2+p^2=n^2+q^2....(1) mn=pq;....(2) now choose m,n,p,q such that the conditions 1,2 satisfied simultaneously....

2. anonymous

are a and b complex numbers ?

3. anonymous

yes... its given..

4. anonymous

how did you get mn = pq

5. anonymous

sorry that will be mn=-pq.

6. anonymous

also you foiled wrong

7. anonymous

(m+ni)^2+(p+qi)^2 = m^2 + 2mni -n^2 + p^2 + 2pqi - q^2 = 0

8. anonymous

so???

9. anonymous

where is 2mni and 2pqi

10. anonymous

2mn=2pq => mn=pq

11. anonymous

in your algebra , i dont see it

12. anonymous

i have done it in single step.

13. anonymous

too fast for me

14. anonymous

sorry im doing another problem, helping student, one sec

15. anonymous

16. anonymous

oh

17. anonymous

2mn i = -2pqi

18. anonymous

ok then mn = -pq, agreed

19. anonymous

something tells me that a,b are integers though originally

20. anonymous

btw: a and b are real

21. anonymous

22. anonymous

our guy here thinks that a and b are complex numbers , i thought this was simpler, lol

23. anonymous

the question is: find a and b (both real) such that (-bi)^2 = a^2

24. anonymous

if a, b are not complex then the relation (a)^2=(-bi)^2 does not hold.

25. anonymous

so a=b=0

26. anonymous

right

27. anonymous

see your question starts with "complex numbers". so i thought that

28. anonymous

sorry

29. anonymous

this equation only have a trivial solution..

30. anonymous

yeah, no fun

31. anonymous

why a = b = 0? is it because a^2 + b^2 = 0 + 0i?

32. anonymous

so the question is more interesting if you allow a and b to be complex , right?

33. anonymous

for real no. only a=b=0 holds

34. anonymous

ok

35. anonymous

a^2 = (-b*i)^2 = b^2*i^2 = b^2 *-1 ,

36. anonymous

can you find any such a,b other than 0 that holds your condition????????????????

37. anonymous

so a^2 = -b^2 , so a^2 + b^2 = 0,

38. anonymous

no because a^2>=0 , and b^2 >= 0

39. anonymous

see i think a,b must be considered as complex otherwise its very easy....

40. anonymous

just solve the damn equation, lol

41. anonymous

if you take them complex then also you will find a solution 0+0i along with some new solution...

42. anonymous

ahhh

43. anonymous

so this equation is equilvalent, find all a,b, such that a^2 + b^2 = 0, for a,b element of C

44. anonymous

damn...i have another typo: -(bi)^2 = a^2...therefore b^2 = a^2 - so the solution is a = b?

45. anonymous

right?

46. anonymous

noooooooooooooooooooooooooooooooooo

47. anonymous

(-bi)^2 does not mean -(bi)^2

48. anonymous

yes i know

49. anonymous

find a and b (both real) such that -(bi)^2 = a^2.......this is the correct problem statement.....sorry....