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