anonymous
  • anonymous
complex numbers: find a, b such that (-bi)^2 = a^2 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • 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....
anonymous
  • anonymous
are a and b complex numbers ?
anonymous
  • anonymous
yes... its given..

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anonymous
  • anonymous
how did you get mn = pq
anonymous
  • anonymous
sorry that will be mn=-pq.
anonymous
  • anonymous
also you foiled wrong
anonymous
  • anonymous
(m+ni)^2+(p+qi)^2 = m^2 + 2mni -n^2 + p^2 + 2pqi - q^2 = 0
anonymous
  • anonymous
so???
anonymous
  • anonymous
where is 2mni and 2pqi
anonymous
  • anonymous
2mn=2pq => mn=pq
anonymous
  • anonymous
in your algebra , i dont see it
anonymous
  • anonymous
i have done it in single step.
anonymous
  • anonymous
too fast for me
anonymous
  • anonymous
sorry im doing another problem, helping student, one sec
anonymous
  • anonymous
i dont follow your logic
anonymous
  • anonymous
oh
anonymous
  • anonymous
2mn i = -2pqi
anonymous
  • anonymous
ok then mn = -pq, agreed
anonymous
  • anonymous
something tells me that a,b are integers though originally
anonymous
  • anonymous
btw: a and b are real
anonymous
  • anonymous
the correct answer is: a=b?
anonymous
  • anonymous
our guy here thinks that a and b are complex numbers , i thought this was simpler, lol
anonymous
  • anonymous
the question is: find a and b (both real) such that (-bi)^2 = a^2
anonymous
  • anonymous
if a, b are not complex then the relation (a)^2=(-bi)^2 does not hold.
anonymous
  • anonymous
so a=b=0
anonymous
  • anonymous
right
anonymous
  • anonymous
see your question starts with "complex numbers". so i thought that
anonymous
  • anonymous
sorry
anonymous
  • anonymous
this equation only have a trivial solution..
anonymous
  • anonymous
yeah, no fun
anonymous
  • anonymous
why a = b = 0? is it because a^2 + b^2 = 0 + 0i?
anonymous
  • anonymous
so the question is more interesting if you allow a and b to be complex , right?
anonymous
  • anonymous
for real no. only a=b=0 holds
anonymous
  • anonymous
ok
anonymous
  • anonymous
a^2 = (-b*i)^2 = b^2*i^2 = b^2 *-1 ,
anonymous
  • anonymous
can you find any such a,b other than 0 that holds your condition????????????????
anonymous
  • anonymous
so a^2 = -b^2 , so a^2 + b^2 = 0,
anonymous
  • anonymous
no because a^2>=0 , and b^2 >= 0
anonymous
  • anonymous
see i think a,b must be considered as complex otherwise its very easy....
anonymous
  • anonymous
just solve the damn equation, lol
anonymous
  • anonymous
if you take them complex then also you will find a solution 0+0i along with some new solution...
anonymous
  • anonymous
ahhh
anonymous
  • anonymous
so this equation is equilvalent, find all a,b, such that a^2 + b^2 = 0, for a,b element of C
anonymous
  • anonymous
damn...i have another typo: -(bi)^2 = a^2...therefore b^2 = a^2 - so the solution is a = b?
anonymous
  • anonymous
right?
anonymous
  • anonymous
noooooooooooooooooooooooooooooooooo
anonymous
  • anonymous
(-bi)^2 does not mean -(bi)^2
anonymous
  • anonymous
yes i know
anonymous
  • anonymous
find a and b (both real) such that -(bi)^2 = a^2.......this is the correct problem statement.....sorry....

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