Simplification imaginary fractionsIn an exercise, a partial fraction expansion has to be done. I have no problem with that, but one of the last steps includes a simplification as follows: ( − 1 2 − 1 6 i ) ( 1 s + 1 + 3 i ) + ( − 1 2 + 1 6 i ) ( 1 s + 1 − 3 i ) = − s + 1 ( s + 1 ) 2 + 3 2 − 1 3 ( 3 ( s + 1 ) 2 + 3 2 ) .My problem is that I do not understand the steps in between these two results. Could anyone explain how this is done? Any help is appreciated!

Question

Simplification imaginary fractionsIn an exercise, a partial fraction expansion has to be done. I have no problem with that, but one of the last steps includes a simplification as follows:      (    −          1      2        −          1      6        i    )        (          1              s        +        1        +        3        i              )    +      (    −          1      2        +          1      6        i    )        (          1              s        +        1        −        3        i              )      =  −            s      +      1              (      s      +      1              )        2            +              3        2              −      1    3        (          3              (        s        +        1                  )          2                +                  3          2                      )    .My problem is that I do not understand the steps in between these two results. Could anyone explain how this is done? Any help is appreciated!

Madelyn Lynch · Accepted Answer

The first step is usually to realify(?) the denominators of your complex fraction. To do this, multiply the first term on the LHS by   1  =            s      +      1      −      3      i              s      +      1      −      3      i       . Multiply the second by   1  =            s      +      1      +      3      i              s      +      1      +      3      i      . These are just the complex conjugates of the denominators. Also, once you obtain the RHS, notice that you can simplify it further. So let&#039;s try:      (    −          1      2        −          1      6        i    )        (          1              s        +        1        +        3        i              )    +      (    −          1      2        +          1      6        i    )        (          1              s        +        1        −        3        i              )      =  −      1    6        (                  3        +        i                    s        +        1        +        3        i              )    +      1    6        (                  −        3        +        i                    s        +        1        −        3        i              )      =  −      1    6        (                  3        +        i                    s        +        1        +        3        i              )        (                  s        +        1        −        3        i                    s        +        1        −        3        i              )    +      1    6        (                  −        3        +        i                    s        +        1        −        3        i              )        (                  s        +        1        +        3        i                    s        +        1        +        3        i              )      =  −      1    6        (                  [        3        (        s        +        1        )        +        3        ]        +        [        −        9        +        (        s        +        1        )        ]        i                    (        s        +        1                  )          2                +                  3          2                      )    +      1    6        (                  [        −        3        (        s        +        1        )        −        3        ]        +        [        −        9        +        (        s        +        1        )        ]        i                    (        s        +        1                  )          2                +                  3          2                      )      =      (                  −        (        s        +        1        )        −        1                    (        s        +        1                  )          2                +                  3          2                      )      =  −      (                  s        +        2                    (        s        +        1                  )          2                +                  3          2                      )

Simplification imaginary fractions In an exercise, a partial fraction expansion has to be done. I h

Answered question

Answer & Explanation

New Questions in Pre-Algebra