Question

complex numbers: find a, b such that (-bi)^2 = a^2 ?

Answer 1

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....

Answer 2

are a and b complex numbers ?

Answer 3

yes...
its given..

Answer 4

how did you get mn = pq

Answer 5

sorry that will be mn=-pq.

Answer 6

also you foiled wrong

Answer 7

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

Answer 8

so???

Answer 9

where is 2mni and 2pqi

Answer 10

2mn=2pq
=> mn=pq

Answer 11

in your algebra , i dont see it

Answer 12

i have done it in single step.

Answer 13

too fast for me

Answer 14

sorry im doing another problem, helping student, one sec

Answer 15

i dont follow your logic

Answer 16

oh

Answer 17

2mn i = -2pqi

Answer 18

ok then mn = -pq, agreed

Answer 19

something tells me that a,b are integers though originally

Answer 20

btw: a and b are real

Answer 21

the correct answer is: a=b?

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

so a=b=0

Answer 26

right

Answer 27

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

Answer 28

sorry

Answer 29

this equation only have a trivial solution..

Answer 30

yeah, no fun

Answer 31

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

Answer 32

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

Answer 33

for real no. only a=b=0 holds

Answer 34

ok

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

just solve the damn equation, lol

Answer 41

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

Answer 42

ahhh

Answer 43

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

Answer 44

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

Answer 45

right?

Answer 46

noooooooooooooooooooooooooooooooooo

Answer 47

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

Answer 48

yes i know

Answer 49

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