SOLUTION: 1.If (a+ib)/(c+id)=A+iB ,then prove that (a-ib)/(c-id)=A-iB
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Question 320383: 1.If (a+ib)/(c+id)=A+iB ,then prove that (a-ib)/(c-id)=A-iB
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
If (a+ib)/(c+id)=A+iB ,then prove that (a-ib)/(c-id)=A-iB:
i = sqrt(-1), i^2 = -1, a and b and c and d and A and B are all real numbers
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(a + ib)/(c + id) (when you divide by a complex number you need to multiply top and bottom by its conjugate, in this case c - id)
(a + ib)/(c + id) * (c - id)/(c - id)
((a + ib)(c - id))/((c + id)(c - id)) (use FOIL on top and bottom)
(ac - iad + ibc - (i^2)bd)/(c^2 - icd + icd - (i^2)d^2)
(ac - iad + ibc - (-1)bd)/(c^2 - icd + icd - (-1)d^2)
(ac + bd - iad + ibc)/(c^2 + d^2)
(ac + bd + i(bc - ad))/(c^2 + d^2)
ac/(c^2 + d^2) + bd/(c^2 + d^2) + i((bc - ad)/(c^2 + d^2))
A = (ac + bd)/(c^2 + d^2)
B = (bc - ad)/(c^2 + d^2)
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(a - ib)/(c - id) (conjugate of c - id is c + id)
(a - ib)/(c - id) * (c + id)/(c + id)
((a - ib)(c + id))/((c - id)(c + id)) (use FOIL on top and bottom)
(ac + iad - ibc - (i^2)bd)/(c^2 + icd - icd - (i^2)d^2)
(ac + iad - ibc - (-1)bd)/(c^2 - (-1)d^2)
(ac + bd - i(bc - ad))/(c^2 + d^2)
ac/(c^2 + d^2) + bd/(c^2 + d^2) - i((bc - ad)/(c^2 + d^2))
A = (ac + bd)/(c^2 + d^2)
B = (bc - ad)/(c^2 + d^2)
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1. (a+ib)/(c+id) = ac/(c^2 + d^2) + bd/(c^2 + d^2) + i((bc - ad)/(c^2 + d^2))
(a+ib)/(c+id)=A+iB
A = (ac + bd)/(c^2 + d^2)
B = (bc - ad)/(c^2 + d^2)
2. (a-ib)/(c-id) = ac/(c^2 + d^2) + bd/(c^2 + d^2) - i((bc - ad)/(c^2 + d^2))
(a-ib)/(c-id)=A-iB
A = (ac + bd)/(c^2 + d^2)
B = (bc - ad)/(c^2 + d^2)
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proven
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