Question

Partial Fractions question:

Answer 1

I realise the below equation is true but how would i have known this was true if normally with partial fractions you add the two terms on the right hand side of the equals.
\[\large \frac{1}{i^{2}+3i+2} = \frac{1}{(i+2)(i+1)} = \frac{1}{i+1} - \frac{1}{i+2}\]

Answer 2

@amistre64

Answer 3

...but this time you minus them

Answer 4

\[ \frac{1}{i^{2}+3i+2} = \frac{1}{(i+2)(i+1)} = \frac{A}{i+1} + \frac{B}{i+2}\]

Answer 5

\[(i+2)(i+1)\times \frac{1}{(i+2)(i+1)} = (i+2)(i+1)\times\left(\frac{A}{i+1} + \frac{B}{i+2}\right)\]

\[\qquad\qquad\qquad\qquad1= (i+2)A+(i+1)B\]

Answer 6

if \( i =-1\)
\[1=A\]


if \(i=-2\)
\[1=-B\]

Answer 7

can you follow that?

Answer 8

yeah....wasn't thinking it through properly. thanks :)

Answer 9

yeah, Unkle did the proper steps which most teachers ive seen omit.

is "i" used here as a normal variable, or is it the complex i ?

Answer 10

not that it matter much, the complex i is treated as a variable until the end when it is simplified