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Can anyone explain to me how (B(1+i)-m)(1+i)-m = B(1+i) squared -m(1+i)-m?

2007-05-07 15:52:43 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

(B(1+i)-m)(1+i) - m =
(B(1+i) - m) + (B(1+i) - m)i - m =
B(1+i) - m + B(1+i)i - mi - m =
B(1+i) + B(1+i)i - m(1+i) - m =
B[ (1+i) + (1+i)i ] - m(1+i) - m =
B[ 1 + i + i + i^2 ] - m(1+i) - m =
B[ 1 + 2i+ i^2 ] - m(1+i) - m =
B(1+i)^2 - m(1+i) - m

2007-05-07 16:02:14 · answer #1 · answered by Anonymous · 0 0

First add m to each side and get (B(1 + i) - m)(1 + i) = B(1 + i)^2 -m(1 + i).

Then multiply (1 + i) to each part of (B(1 + i) - m) and write it out: B(1 + i)(1 + i) -m(1 + i). Combine the (1 + i)(1 + i) part into (1 + i)^2, and you get exactly what's on the right side.

2007-05-07 23:04:24 · answer #2 · answered by lynxdaemon19 2 · 0 0

1) Please tell us what field you are in (e.g., does i represent the square root of -1?)
2) rewrite the equation using * for multiplication and ^ for powers, such as B*(1+i)^2

2007-05-07 23:02:59 · answer #3 · answered by Raymond 7 · 0 0

b=i c= another one a d=neither . good luck with the answers .

2007-05-07 22:58:07 · answer #4 · answered by Anonymous · 0 1

nope sorry thats over my head

2007-05-07 22:56:19 · answer #5 · answered by JENNIFER W 2 · 0 1

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