What is ((2+i)-(3-4i))/((5-i)(3+i)) ?

The solution to ((2+i)-(3-4i))/((5-i)(3+i)) is -3/130+41/130 i

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simplify ((2+i)-(3-4i))/((5-i)(3+i))

simplify

\frac{( 2 + i ) - ( 3 - 4 i )}{( 5 - i ) ( 3 + i )}

- \frac{3}{130} + \frac{41}{130} i

Solution steps

\frac{( 2 + i ) - ( 3 - 4 i )}{( 5 - i ) ( 3 + i )}

Apply rule:

(a) = a

(2 + i) = 2 + i

= \frac{2 + i - ( 3 - 4 i )}{( 5 - i ) ( 3 + i )}

Simplify

2 + i - (3 - 4 i) : 5 i - 1

= \frac{5 i - 1}{( 5 - i ) ( 3 + i )}

(5 - i) (3 + i) = 16 + 2 i

= \frac{5 i - 1}{16 + 2 i}

Rewrite

5 i - 1

in standard complex form:

- 1 + 5 i

= \frac{- 1 + 5 i}{16 + 2 i}

Multiply by the conjugate

\frac{16 - 2 i}{16 - 2 i}

= \frac{( - 1 + 5 i ) ( 16 - 2 i )}{( 16 + 2 i ) ( 16 - 2 i )}

Simplify

\frac{( - 1 + 5 i ) ( 16 - 2 i )}{( 16 + 2 i ) ( 16 - 2 i )} : \frac{- 6 + 82 i}{260}

= \frac{- 6 + 82 i}{260}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{- 6}{260} + \frac{82}{260} i

Simplify

\frac{- 6}{260} + \frac{82}{260} i : - \frac{3}{130} + \frac{41}{130} i

= - \frac{3}{130} + \frac{41}{130} i

\frac{5 a ^{- 4}}{2 c}

\frac{4 ^{4}}{4}

12 x + x^{2} = - 7

4 x^{2} - 12 x - 16

What is ((2+i)-(3-4i))/((5-i)(3+i)) ?
The solution to ((2+i)-(3-4i))/((5-i)(3+i)) is -3/130+41/130 i